找出第n个数字,其二进制表示形式是回文。在考虑二进制表示时,请不要考虑前导零。将二进制表示为回文的第一个数字视为1,而不是0
例子:
Input : 1
Output : 1
1st Number whose binary representation
is palindrome is 1 (1)
Input : 9
Output : 27
9th Number whose binary representation
is palindrome is 27 (11011)
方法1:天真
如果天数是回文数,则幼稚的方法是遍历1到2 ^ 31 – 1的所有整数并增加回文数。当回文数达到所需的n时,中断循环并返回当前整数。
C++
// C++ program to find n-th number whose binary
// representation is palindrome.
#include
using namespace std;
// Finds if the kth bit is set in the binary
// representation
int isKthBitSet(int x, int k)
{
return (x & (1 << (k - 1))) ? 1 : 0;
}
// Returns the position of leftmost set bit
// in the binary representation
int leftmostSetBit(int x)
{
int count = 0;
while (x) {
count++;
x = x >> 1;
}
return count;
}
// Finds whether the integer in binary
// representation is palindrome or not
int isBinPalindrome(int x)
{
int l = leftmostSetBit(x);
int r = 1;
// One by one compare bits
while (l > r) {
// Compare left and right bits and converge
if (isKthBitSet(x, l) != isKthBitSet(x, r))
return 0;
l--;
r++;
}
return 1;
}
int findNthPalindrome(int n)
{
int pal_count = 0;
// Start from 1, traverse through
// all the integers
int i = 0;
for (i = 1; i <= INT_MAX; i++) {
if (isBinPalindrome(i)) {
pal_count++;
}
// If we reach n, break the loop
if (pal_count == n)
break;
}
return i;
}
// Driver code
int main()
{
int n = 9;
// Function Call
cout << findNthPalindrome(n);
}
// This code is contributed
// by Akanksha Rai
C
// C program to find n-th number whose binary
// representation is palindrome.
#include
#define INT_MAX 2147483647
// Finds if the kth bit is set in the binary
// representation
int isKthBitSet(int x, int k)
{
return (x & (1 << (k - 1))) ? 1 : 0;
}
// Returns the position of leftmost set bit
// in the binary representation
int leftmostSetBit(int x)
{
int count = 0;
while (x) {
count++;
x = x >> 1;
}
return count;
}
// Finds whether the integer in binary
// representation is palindrome or not
int isBinPalindrome(int x)
{
int l = leftmostSetBit(x);
int r = 1;
// One by one compare bits
while (l > r) {
// Compare left and right bits and converge
if (isKthBitSet(x, l) != isKthBitSet(x, r))
return 0;
l--;
r++;
}
return 1;
}
int findNthPalindrome(int n)
{
int pal_count = 0;
// Start from 1, traverse through
// all the integers
int i = 0;
for (i = 1; i <= INT_MAX; i++) {
if (isBinPalindrome(i)) {
pal_count++;
}
// If we reach n, break the loop
if (pal_count == n)
break;
}
return i;
}
// Driver code
int main()
{
int n = 9;
// Function Call
printf("%d", findNthPalindrome(n));
}
Java
// Java program to find n-th
// number whose binary
// representation is palindrome.
import java.io.*;
class GFG {
static int INT_MAX = 2147483647;
// Finds if the kth bit
// is set in the binary
// representation
static int isKthBitSet(int x, int k)
{
return ((x & (1 << (k - 1))) > 0) ? 1 : 0;
}
// Returns the position of
// leftmost set bit in the
// binary representation
static int leftmostSetBit(int x)
{
int count = 0;
while (x > 0) {
count++;
x = x >> 1;
}
return count;
}
// Finds whether the integer
// in binary representation is
// palindrome or not
static int isBinPalindrome(int x)
{
int l = leftmostSetBit(x);
int r = 1;
// One by one compare bits
while (l > r) {
// Compare left and right
// bits and converge
if (isKthBitSet(x, l) != isKthBitSet(x, r))
return 0;
l--;
r++;
}
return 1;
}
static int findNthPalindrome(int n)
{
int pal_count = 0;
// Start from 1, traverse
// through all the integers
int i = 0;
for (i = 1; i <= INT_MAX; i++) {
if (isBinPalindrome(i) > 0) {
pal_count++;
}
// If we reach n,
// break the loop
if (pal_count == n)
break;
}
return i;
}
// Driver code
public static void main(String[] args)
{
int n = 9;
// Function Call
System.out.println(findNthPalindrome(n));
}
}
// This code is contributed
// by anuj_67.
Python3
# Python 3 program to find n-th number
# whose binary representation is palindrome.
INT_MAX = 2147483647
# Finds if the kth bit is set in
# the binary representation
def isKthBitSet(x, k):
return 1 if (x & (1 << (k - 1))) else 0
# Returns the position of leftmost
# set bit in the binary representation
def leftmostSetBit(x):
count = 0
while (x):
count += 1
x = x >> 1
return count
# Finds whether the integer in binary
# representation is palindrome or not
def isBinPalindrome(x):
l = leftmostSetBit(x)
r = 1
# One by one compare bits
while (l > r):
# Compare left and right bits
# and converge
if (isKthBitSet(x, l) != isKthBitSet(x, r)):
return 0
l -= 1
r += 1
return 1
def findNthPalindrome(n):
pal_count = 0
# Start from 1, traverse
# through all the integers
i = 0
for i in range(1, INT_MAX + 1):
if (isBinPalindrome(i)):
pal_count += 1
# If we reach n, break the loop
if (pal_count == n):
break
return i
# Driver code
if __name__ == "__main__":
n = 9
# Function Call
print(findNthPalindrome(n))
# This code is contributed
# by ChitraNayal
C#
// C# program to find n-th
// number whose binary
// representation is palindrome.
using System;
class GFG {
static int INT_MAX = 2147483647;
// Finds if the kth bit
// is set in the binary
// representation
static int isKthBitSet(int x, int k)
{
return ((x & (1 << (k - 1))) > 0) ? 1 : 0;
}
// Returns the position of
// leftmost set bit in the
// binary representation
static int leftmostSetBit(int x)
{
int count = 0;
while (x > 0) {
count++;
x = x >> 1;
}
return count;
}
// Finds whether the integer
// in binary representation is
// palindrome or not
static int isBinPalindrome(int x)
{
int l = leftmostSetBit(x);
int r = 1;
// One by one compare bits
while (l > r) {
// Compare left and right
// bits and converge
if (isKthBitSet(x, l) != isKthBitSet(x, r))
return 0;
l--;
r++;
}
return 1;
}
static int findNthPalindrome(int n)
{
int pal_count = 0;
// Start from 1, traverse
// through all the integers
int i = 0;
for (i = 1; i <= INT_MAX; i++) {
if (isBinPalindrome(i) > 0) {
pal_count++;
}
// If we reach n,
// break the loop
if (pal_count == n)
break;
}
return i;
}
// Driver code
static public void Main()
{
int n = 9;
// Function Call
Console.WriteLine(findNthPalindrome(n));
}
}
// This code is contributed ajit
PHP
> 1;
}
return $count;
}
// Finds whether the integer in binary
// representation is palindrome or not
function isBinPalindrome($x)
{
$l = leftmostSetBit($x);
$r = 1;
// One by one compare bits
while ($l > $r)
{
// Compare left and right bits
// and converge
if (isKthBitSet($x, $l) !=
isKthBitSet($x, $r))
return 0;
$l--;
$r++;
}
return 1;
}
function findNthPalindrome($n)
{
$pal_count = 0;
// Start from 1, traverse through
// all the integers
$i = 0;
for ($i = 1; $i <= PHP_INT_MAX; $i++)
{
if (isBinPalindrome($i))
{
$pal_count++;
}
// If we reach n, break the loop
if ($pal_count == $n)
break;
}
return $i;
}
// Driver code
$n = 9;
// Function Call
echo (findNthPalindrome($n));
// This code is contributed by jit_t
?>
Javascript
C++
// C++ program to find n-th palindrome
#include
using namespace std;
// utility function which is used to
// convert binary string into integer
int convertStringToInt(string s)
{
int num = 0;
// convert binary string into integer
for (int i = 0; i < s.size(); i++) {
num = num * 2 + (s[i] - '0');
}
return num;
}
// function to find nth binary palindrome number
int getNthNumber(int n)
{
// base case
if (n == 1)
return 1;
n--;
// stores the binary palindrome string
queue q;
// add 2nd binary palindrome string
q.push("11");
// runs till the nth binary palindrome number
while (!q.empty()) {
// remove curr binary palindrome string from queue
string curr = q.front();
q.pop();
n--;
// if n==0 then we find the n'th binary palindrome
// so we return our answer
if (!n) {
return convertStringToInt(curr);
}
int mid = curr.size() / 2;
// if length is even .so it is our first case
// we have two choices
if (curr.size() % 2 == 0) {
string s0 = curr, s1 = curr;
s0.insert(mid, "0");
s1.insert(mid, "1");
q.push(s0);
q.push(s1);
}
// if length is odd .so it is our second case
// we have only one choice
else {
string ch(1, curr[mid]);
string temp = curr;
temp.insert(mid, ch);
q.push(temp);
}
}
return 0;
}
// Driver Code
int main()
{
int n = 9;
// Function Call
cout << getNthNumber(n);
}
// This code is contributed by Sagar Jangra and Naresh
// Saharan
Java
// Java program to find n-th palindrome
import java.io.*;
import java.util.*;
class GFG {
// utility function which is used to
// convert binary string into integer
public static int convertStringToInt(String s)
{
int ans = 0;
// convert binary string into integer
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '1')
ans += 1 << i;
}
return ans;
}
// function to find nth binary palindrome number
public static int getNthNumber(int n)
{
// stores the binary palindrome string
Queue q = new LinkedList<>();
// base case
if (n == 1)
return 1;
n = n - 1;
// add 2nd binary palindrome string
q.add("11");
// runs till the nth binary palindrome number
while (n-- > 0) {
// remove curr binary palindrome string from
// queue
String curr = q.remove();
// if n==0 then we find the n'th binary
// palindrome so we return our answer
if (n == 0)
return convertStringToInt(curr);
// calculate length of curr binary palindrome
// string
int len = curr.length();
// if length is even .so it is our first case
// we have two choices
if (len % 2 == 0) {
q.add(curr.substring(0, len / 2) + "0"
+ curr.substring(len / 2));
q.add(curr.substring(0, len / 2) + "1"
+ curr.substring(len / 2));
}
// if length is odd .so it is our second case
// we have only one choice
else {
char midChar = curr.charAt(len / 2);
q.add(curr.substring(0, len / 2) + midChar
+ curr.substring(len / 2));
}
}
return -1;
}
// Driver code
public static void main(String[] args)
{
int n = 9;
// Function Call
System.out.println(getNthNumber(n));
}
}
// This code is contributed by Naresh Saharan and Sagar
// Jangra
Python3
# Python program to find n-th palindrome
# utility function which is used to
# convert binary string into integer
def convertStringToInt(s):
ans = 0
# convert binary string into integer
for i in range(len(s)):
ans = ans * 2 + (ord(s[i]) - ord('0'))
return ans
# function to find nth binary palindrome number
def getNthNumber(n):
# stores the binary palindrome string
q = []
# base case
if(n == 1):
return 1
n = n - 1
# add 2nd binary palindrome string
q.append("11")
# runs till the nth binary palindrome number
while(len(q) != 0):
# remove curr binary palindrome string from
# queue
curr = q.pop(0)
n -= 1
# if n==0 then we find the n'th binary
# palindrome so we return our answer
if(n ==0):
return convertStringToInt(curr)
# calculate length of curr binary palindrome
# string
lenn = len(curr)
# if length is even .so it is our first case
# we have two choices
if (len(curr) % 2 == 0):
q.append(curr[0:int(lenn/2)]+"0"+curr[int(lenn/2):])
q.append(curr[0:int(lenn/2)]+"1"+curr[int(lenn/2):])
# if length is odd .so it is our second case
# we have only one choice
else:
midChar = curr[int(lenn/2)]
q.append(curr[0:int(lenn/2)]+midChar+curr[int(lenn/2):])
return 0
# Driver code
n = 9
# Function Call
print(getNthNumber(n))
# This code is contributed by avanitrachhadiya2155
C
// Efficient C program to find n-th palindrome
#include
#define INT_SIZE 32
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
int constructNthNumber(int group_no, int aux_num, int op)
{
int a[INT_SIZE] = { 0 };
int num = 0, len_f;
int i = 0;
// No need to insert any bit in the middle
if (op == 2) {
// Length of the final binary representation
len_f = 2 * group_no;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
// Start filling the a[] from middle,
// with the aux_num binary representation
while (aux_num) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Insert bit 0 in the middle
else if (op == 0) {
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 0;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num) {
// Get the auxiliary number's ith bit and fill
// around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 1;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num) {
// Get the auxiliary number's ith bit and fill
// around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Convert the number to decimal from binary
for (i = 0; i < len_f; i++)
num += (1 << i) * a[i];
return num;
}
// Will return the nth binary palindrome number
int getNthNumber(int n)
{
int group_no = 0, group_offset;
int count_upto_group = 0, count_temp = 1;
int op, aux_num;
// Add number of elements in all the groups,
// until the group of the nth number is found
while (count_temp < n) {
group_no++;
// Total number of elements until this group
count_upto_group = count_temp;
count_temp += 3 * (1 << (group_no - 1));
}
// Element's offset position in the group
group_offset = n - count_upto_group - 1;
// Finding which bit to be placed in the
// middle and finding the number, which we
// will fill from the middle in both
// directions
if ((group_offset + 1) <= (1 << (group_no - 1))) {
op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
aux_num = group_offset;
}
else {
if (((group_offset + 1)
- (1 << (group_no - 1))) % 2)
op = 0; // Need to Insert 0 at middle
else
op = 1; // Need to Insert 1 at middle
aux_num
= ((group_offset) - (1 << (group_no - 1))) / 2;
}
return constructNthNumber(group_no, aux_num, op);
}
// Driver code
int main()
{
int n = 9;
// Function Call
printf("%d", getNthNumber(n));
return 0;
}
Java
// Efficient Java program to find n-th palindrome
class GFG {
static int INT_SIZE = 32;
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
static int constructNthNumber(int group_no, int aux_num,
int op)
{
int a[] = new int[INT_SIZE];
int num = 0, len_f;
int i = 0;
// No need to insert any bit in the middle
if (op == 2) {
// Length of the final binary representation
len_f = 2 * group_no;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
// Start filling the a[] from middle,
// with the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Insert bit 0 in the middle
else if (op == 0) {
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 0;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 1;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Convert the number to decimal from binary
for (i = 0; i < len_f; i++)
num += (1 << i) * a[i];
return num;
}
// Will return the nth binary palindrome number
static int getNthNumber(int n)
{
int group_no = 0, group_offset;
int count_upto_group = 0, count_temp = 1;
int op, aux_num;
// Add number of elements in all the groups,
// until the group of the nth number is found
while (count_temp < n) {
group_no++;
// Total number of elements until this group
count_upto_group = count_temp;
count_temp += 3 * (1 << (group_no - 1));
}
// Element's offset position in the group
group_offset = n - count_upto_group - 1;
// Finding which bit to be placed in the
// middle and finding the number, which we
// will fill from the middle in both
// directions
if ((group_offset + 1) <= (1 << (group_no - 1))) {
op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
aux_num = group_offset;
}
else {
if (((group_offset + 1)
- (1 << (group_no - 1))) % 2 == 1)
op = 0; // Need to Insert 0 at middle
else
op = 1; // Need to Insert 1 at middle
aux_num
= ((group_offset)
- (1 << (group_no - 1))) / 2;
}
return constructNthNumber(group_no, aux_num, op);
}
// Driver code
public static void main(String[] args)
{
int n = 9;
// Function Call
System.out.printf("%d", getNthNumber(n));
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Efficient Python3 program to find n-th palindrome
INT_SIZE = 32
# Construct the nth binary palindrome with the
# given group number, aux_number and operation type
def constructNthNumber(group_no, aux_num, op):
a = [0] * INT_SIZE
num, i = 0, 0
# No need to insert any bit in the middle
if op == 2:
# Length of the final binary representation
len_f = 2 * group_no
# Fill first and last bit as 1
a[len_f - 1] = a[0] = 1
# Start filling the a[] from middle,
# with the aux_num binary representation
while aux_num:
# Get the auxiliary number's ith
# bit and fill around middle
a[group_no + i] = a[group_no - 1 - i] = \
aux_num & 1
aux_num = aux_num >> 1
i += 1
# Insert bit 0 in the middle
elif op == 0:
# Length of the final binary representation
len_f = 2 * group_no + 1
# Fill first and last bit as 1
a[len_f - 1] = a[0] = 1
a[group_no] = 0
# Start filling the a[] from middle, with
# the aux_num binary representation
while aux_num:
# Get the auxiliary number's ith
# bit and fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i] = \
aux_num & 1
aux_num = aux_num >> 1
i += 1
else: # Insert bit 1 in the middle
# Length of the final binary representation
len_f = 2 * group_no + 1
# Fill first and last bit as 1
a[len_f - 1] = a[0] = 1
a[group_no] = 1
# Start filling the a[] from middle, with
# the aux_num binary representation
while aux_num:
# Get the auxiliary number's ith
# bit and fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i] = \
aux_num & 1
aux_num = aux_num >> 1
i += 1
# Convert the number to decimal from binary
for i in range(0, len_f):
num += (1 << i) * a[i]
return num
# Will return the nth binary palindrome number
def getNthNumber(n):
group_no = 0
count_upto_group, count_temp = 0, 1
# Add number of elements in all the groups,
# until the group of the nth number is found
while count_temp < n:
group_no += 1
# Total number of elements until this group
count_upto_group = count_temp
count_temp += 3 * (1 << (group_no - 1))
# Element's offset position in the group
group_offset = n - count_upto_group - 1
# Finding which bit to be placed in the
# middle and finding the number, which we
# will fill from the middle in both directions
if (group_offset + 1) <= (1 << (group_no - 1)):
op = 2 # No need to put extra bit in middle
# We need to fill this auxiliary number
# in binary form the middle in both directions
aux_num = group_offset
else:
if (((group_offset + 1) -
(1 << (group_no - 1))) % 2):
op = 0 # Need to Insert 0 at middle
else:
op = 1 # Need to Insert 1 at middle
aux_num = (((group_offset) -
(1 << (group_no - 1))) // 2)
return constructNthNumber(group_no, aux_num, op)
# Driver code
if __name__ == "__main__":
n = 9
# Function Call
print(getNthNumber(n))
# This code is contributed by Rituraj Jain
C#
// Efficient C# program to find n-th palindrome
using System;
class GFG {
static int INT_SIZE = 32;
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
static int constructNthNumber(int group_no, int aux_num,
int op)
{
int[] a = new int[INT_SIZE];
int num = 0, len_f;
int i = 0;
// No need to insert any bit in the middle
if (op == 2) {
// Length of the final binary representation
len_f = 2 * group_no;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
// Start filling the a[] from middle,
// with the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + i] = a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Insert bit 0 in the middle
else if (op == 0) {
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 0;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 1;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Convert the number to decimal from binary
for (i = 0; i < len_f; i++)
num += (1 << i) * a[i];
return num;
}
// Will return the nth binary palindrome number
static int getNthNumber(int n)
{
int group_no = 0, group_offset;
int count_upto_group = 0, count_temp = 1;
int op, aux_num;
// Add number of elements in all the groups,
// until the group of the nth number is found
while (count_temp < n) {
group_no++;
// Total number of elements until this group
count_upto_group = count_temp;
count_temp += 3 * (1 << (group_no - 1));
}
// Element's offset position in the group
group_offset = n - count_upto_group - 1;
// Finding which bit to be placed in the
// middle and finding the number, which we
// will fill from the middle in both
// directions
if ((group_offset + 1) <= (1 << (group_no - 1))) {
op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
aux_num = group_offset;
}
else {
if (((group_offset + 1) - (1 << (group_no - 1)))
% 2
== 1)
op = 0; // Need to Insert 0 at middle
else
op = 1; // Need to Insert 1 at middle
aux_num
= ((group_offset) - (1 << (group_no - 1)))
/ 2;
}
return constructNthNumber(group_no, aux_num, op);
}
// Driver code
public static void Main(String[] args)
{
int n = 9;
// Function Call
Console.Write("{0}", getNthNumber(n));
}
}
// This code contributed by Rajput-Ji
PHP
> 1;
$i++;
}
}
// Insert bit 0 in the middle
else if ($op == 0)
{
// Length of the final binary representation
$len_f = 2 * $group_no + 1;
// Fill first and last bit as 1
$a[$len_f - 1] = $a[0] = 1;
$a[$group_no] = 0;
/* Start filling the a[] from middle, with
the aux_num binary representation */
while ($aux_num)
{
// Get the auxiliary number's ith bit and fill
// around middle
$a[$group_no + 1 + $i] = $a[$group_no - 1 - $i]
= $aux_num & 1;
$aux_num = $aux_num >> 1;
$i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
$len_f = 2 * $group_no + 1;
// Fill first and last bit as 1
$a[$len_f - 1] = $a[0] = 1;
$a[$group_no] = 1;
/* Start filling the a[] from middle, with
the aux_num binary representation */
while ($aux_num)
{
// Get the auxiliary number's ith bit and fill
// around middle
$a[$group_no + 1 + $i] = $a[$group_no - 1 - $i]
= $aux_num & 1;
$aux_num = $aux_num >> 1;
$i++;
}
}
/* Convert the number to decimal from binary */
for ($i = 0; $i < $len_f; $i++)
$num += (1 << $i) * $a[$i];
return $num;
}
/* Will return the nth binary palindrome number */
function getNthNumber($n)
{
$group_no = 0;
$count_upto_group = 0;
$count_temp = 1;
$op=$aux_num=0;
/* Add number of elements in all the groups,
until the group of the nth number is found */
while ($count_temp < $n)
{
$group_no++;
// Total number of elements until this group
$count_upto_group = $count_temp;
$count_temp += 3 * (1 << ($group_no - 1));
}
// Element's offset position in the group
$group_offset = $n - $count_upto_group - 1;
/* Finding which bit to be placed in the
middle and finding the number, which we
will fill from the middle in both
directions */
if (($group_offset + 1) <= (1 << ($group_no - 1)))
{
$op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
$aux_num = $group_offset;
}
else
{
if ((($group_offset + 1) - (1 << ($group_no - 1))) % 2)
$op = 0; // Need to Insert 0 at middle
else
$op = 1; // Need to Insert 1 at middle
$aux_num = (int)((($group_offset) - (1 << ($group_no - 1))) / 2);
}
return constructNthNumber($group_no, $aux_num, $op);
}
// Driver code
$n = 9;
// Function Call
print(getNthNumber($n));
// This code is contributed by mits
?>
输出
27
此解决方案的时间复杂度为O(x) ,其中x是结果数。
请注意,x的值通常比n大得多。
方法2:使用BFS
首先,采用这种方法,我们只需将此字符串添加“ 11”到队列中即可。然后,对于每个字符串,我们有两种情况。 IE
- 如果偶数长度的CURR字符串然后在CURR字符串的中间加“0”和“1”,并把它添加到队列。
- 如果curr字符串的长度为奇数,则将curr字符串的中字符添加到结果字符串,然后将其添加到队列中。
下面是上述方法的实现:
C++
// C++ program to find n-th palindrome
#include
using namespace std;
// utility function which is used to
// convert binary string into integer
int convertStringToInt(string s)
{
int num = 0;
// convert binary string into integer
for (int i = 0; i < s.size(); i++) {
num = num * 2 + (s[i] - '0');
}
return num;
}
// function to find nth binary palindrome number
int getNthNumber(int n)
{
// base case
if (n == 1)
return 1;
n--;
// stores the binary palindrome string
queue q;
// add 2nd binary palindrome string
q.push("11");
// runs till the nth binary palindrome number
while (!q.empty()) {
// remove curr binary palindrome string from queue
string curr = q.front();
q.pop();
n--;
// if n==0 then we find the n'th binary palindrome
// so we return our answer
if (!n) {
return convertStringToInt(curr);
}
int mid = curr.size() / 2;
// if length is even .so it is our first case
// we have two choices
if (curr.size() % 2 == 0) {
string s0 = curr, s1 = curr;
s0.insert(mid, "0");
s1.insert(mid, "1");
q.push(s0);
q.push(s1);
}
// if length is odd .so it is our second case
// we have only one choice
else {
string ch(1, curr[mid]);
string temp = curr;
temp.insert(mid, ch);
q.push(temp);
}
}
return 0;
}
// Driver Code
int main()
{
int n = 9;
// Function Call
cout << getNthNumber(n);
}
// This code is contributed by Sagar Jangra and Naresh
// Saharan
Java
// Java program to find n-th palindrome
import java.io.*;
import java.util.*;
class GFG {
// utility function which is used to
// convert binary string into integer
public static int convertStringToInt(String s)
{
int ans = 0;
// convert binary string into integer
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '1')
ans += 1 << i;
}
return ans;
}
// function to find nth binary palindrome number
public static int getNthNumber(int n)
{
// stores the binary palindrome string
Queue q = new LinkedList<>();
// base case
if (n == 1)
return 1;
n = n - 1;
// add 2nd binary palindrome string
q.add("11");
// runs till the nth binary palindrome number
while (n-- > 0) {
// remove curr binary palindrome string from
// queue
String curr = q.remove();
// if n==0 then we find the n'th binary
// palindrome so we return our answer
if (n == 0)
return convertStringToInt(curr);
// calculate length of curr binary palindrome
// string
int len = curr.length();
// if length is even .so it is our first case
// we have two choices
if (len % 2 == 0) {
q.add(curr.substring(0, len / 2) + "0"
+ curr.substring(len / 2));
q.add(curr.substring(0, len / 2) + "1"
+ curr.substring(len / 2));
}
// if length is odd .so it is our second case
// we have only one choice
else {
char midChar = curr.charAt(len / 2);
q.add(curr.substring(0, len / 2) + midChar
+ curr.substring(len / 2));
}
}
return -1;
}
// Driver code
public static void main(String[] args)
{
int n = 9;
// Function Call
System.out.println(getNthNumber(n));
}
}
// This code is contributed by Naresh Saharan and Sagar
// Jangra
Python3
# Python program to find n-th palindrome
# utility function which is used to
# convert binary string into integer
def convertStringToInt(s):
ans = 0
# convert binary string into integer
for i in range(len(s)):
ans = ans * 2 + (ord(s[i]) - ord('0'))
return ans
# function to find nth binary palindrome number
def getNthNumber(n):
# stores the binary palindrome string
q = []
# base case
if(n == 1):
return 1
n = n - 1
# add 2nd binary palindrome string
q.append("11")
# runs till the nth binary palindrome number
while(len(q) != 0):
# remove curr binary palindrome string from
# queue
curr = q.pop(0)
n -= 1
# if n==0 then we find the n'th binary
# palindrome so we return our answer
if(n ==0):
return convertStringToInt(curr)
# calculate length of curr binary palindrome
# string
lenn = len(curr)
# if length is even .so it is our first case
# we have two choices
if (len(curr) % 2 == 0):
q.append(curr[0:int(lenn/2)]+"0"+curr[int(lenn/2):])
q.append(curr[0:int(lenn/2)]+"1"+curr[int(lenn/2):])
# if length is odd .so it is our second case
# we have only one choice
else:
midChar = curr[int(lenn/2)]
q.append(curr[0:int(lenn/2)]+midChar+curr[int(lenn/2):])
return 0
# Driver code
n = 9
# Function Call
print(getNthNumber(n))
# This code is contributed by avanitrachhadiya2155
输出
27
时间复杂度: O(N)
空间复杂度: O(N)
方法3:构建第n个回文
我们可以使用以下方法直接以其二进制表示形式构造第n个二进制回文。
如果我们观察到头几个二进制回文
* | nth Binary |
n | Palindrome | Group
| |
--------------------------- Group 0
1 ---> 1 (1)
Group 1 (Will have binary representation of length 2*(1)
and 2*(1) + 1)
Fix the first and last bit as 1 and insert nothing
(|) in between. Length is 2*(1)
2 ---> 1|1 (3)
Fix the first and last bit as 1 and insert bit 0
in between. Length is 2*(1) + 1
3 ---> 101 (5)
Fix the first and last bit as 1 and insert bit 1
in between. Length is 2*(1) + 1
4 ---> 111 (7)
F
Group 2 (Will have binary representation of length
2*(2) and 2*(2) + 1). Fix the first and last
bit as 1 and insert nothing (|) at middle.
And put 0 in binary format in both directions
from middle. Length is 2*(2)
5 ---> 10|01
Fix the first and last bit as 1 and insert
nothing (|) at middle. And put 1 in binary
format in both directions from middle.
Length is 2*(2)
6 ---> 11|11
7 ---> 10001
8 ---> 10101
9 ---> 11011
10 ---> 11111
Group 3 (Will have binary representation of
length 2*(3) and 2*(3) + 1)
11 ---> 100|001
12 ---> 101|101
13 ---> 110|011
14 ---> 111|111
15 ---> 1000001
16 ---> 1001001
17 ---> 1010101
18 ---> 1011101
19 ---> 1100011
20 ---> 1101011
21 ---> 1110111
22 ---> 1111111
--------------------
算法:
1)我们可以将回文数集划分为几组。
2)第n组将具有(2 ^(n-1)+ 2 ^ n = 3 * 2 ^(n-1))个二元回文数
3)使用给定的数字,我们可以找到它所属的组以及该组中的偏移量。
4)由于不考虑前导零,因此我们应使用位1作为二进制表示形式中数字的开始位和结束位
5)然后我们将根据groupno和groupoffset填充其他位
6)根据偏移量,我们可以找到应该在中间插入哪个位(|(nothing)或0或1),然后
应从中间向两个方向放置哪个数字(二进制形式)(1或2或3或4或..)
考虑下面的例子
Let us construct the 8th binary palindrome number
The group number will be 2, and no.of elements
before that group are 1 + 3 * 2^1 which is 4
So the offset for the 8th element will be 8 - 4
- 1 = 3
And first 2^(groupno - 1) = 2^1, elements will
have even length(in binary representation) of
2*groupno, next 2^groupno elements will have odd
length(in binary representation) of 2*groupno + 1
Place bit 1 as the first bit and as the last bit
(firstbit: 0, last bit: 2*groupno or 2*groupno - 1)
As the offset is 3, 4th(3 + 1) element in the
group, will have odd length and have 1 in the
middle
Below is the table of middle bit to be used for
the given offset for the group 2
offset middle bit
0 |
1 |
2 0
3 1
4 0
5 1
And we should be filling the binary representation
of number 0(((groupoffset) - 2^(groupno-1)) /2)
from middle n both directions
1 0 1 0 1
FirstElement Number MiddleElement Number LastElement
1 0 1 0 1
The 8th number will be 21
下面是上述想法的实现:
C
// Efficient C program to find n-th palindrome
#include
#define INT_SIZE 32
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
int constructNthNumber(int group_no, int aux_num, int op)
{
int a[INT_SIZE] = { 0 };
int num = 0, len_f;
int i = 0;
// No need to insert any bit in the middle
if (op == 2) {
// Length of the final binary representation
len_f = 2 * group_no;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
// Start filling the a[] from middle,
// with the aux_num binary representation
while (aux_num) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Insert bit 0 in the middle
else if (op == 0) {
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 0;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num) {
// Get the auxiliary number's ith bit and fill
// around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 1;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num) {
// Get the auxiliary number's ith bit and fill
// around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Convert the number to decimal from binary
for (i = 0; i < len_f; i++)
num += (1 << i) * a[i];
return num;
}
// Will return the nth binary palindrome number
int getNthNumber(int n)
{
int group_no = 0, group_offset;
int count_upto_group = 0, count_temp = 1;
int op, aux_num;
// Add number of elements in all the groups,
// until the group of the nth number is found
while (count_temp < n) {
group_no++;
// Total number of elements until this group
count_upto_group = count_temp;
count_temp += 3 * (1 << (group_no - 1));
}
// Element's offset position in the group
group_offset = n - count_upto_group - 1;
// Finding which bit to be placed in the
// middle and finding the number, which we
// will fill from the middle in both
// directions
if ((group_offset + 1) <= (1 << (group_no - 1))) {
op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
aux_num = group_offset;
}
else {
if (((group_offset + 1)
- (1 << (group_no - 1))) % 2)
op = 0; // Need to Insert 0 at middle
else
op = 1; // Need to Insert 1 at middle
aux_num
= ((group_offset) - (1 << (group_no - 1))) / 2;
}
return constructNthNumber(group_no, aux_num, op);
}
// Driver code
int main()
{
int n = 9;
// Function Call
printf("%d", getNthNumber(n));
return 0;
}
Java
// Efficient Java program to find n-th palindrome
class GFG {
static int INT_SIZE = 32;
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
static int constructNthNumber(int group_no, int aux_num,
int op)
{
int a[] = new int[INT_SIZE];
int num = 0, len_f;
int i = 0;
// No need to insert any bit in the middle
if (op == 2) {
// Length of the final binary representation
len_f = 2 * group_no;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
// Start filling the a[] from middle,
// with the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Insert bit 0 in the middle
else if (op == 0) {
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 0;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 1;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i]
= a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Convert the number to decimal from binary
for (i = 0; i < len_f; i++)
num += (1 << i) * a[i];
return num;
}
// Will return the nth binary palindrome number
static int getNthNumber(int n)
{
int group_no = 0, group_offset;
int count_upto_group = 0, count_temp = 1;
int op, aux_num;
// Add number of elements in all the groups,
// until the group of the nth number is found
while (count_temp < n) {
group_no++;
// Total number of elements until this group
count_upto_group = count_temp;
count_temp += 3 * (1 << (group_no - 1));
}
// Element's offset position in the group
group_offset = n - count_upto_group - 1;
// Finding which bit to be placed in the
// middle and finding the number, which we
// will fill from the middle in both
// directions
if ((group_offset + 1) <= (1 << (group_no - 1))) {
op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
aux_num = group_offset;
}
else {
if (((group_offset + 1)
- (1 << (group_no - 1))) % 2 == 1)
op = 0; // Need to Insert 0 at middle
else
op = 1; // Need to Insert 1 at middle
aux_num
= ((group_offset)
- (1 << (group_no - 1))) / 2;
}
return constructNthNumber(group_no, aux_num, op);
}
// Driver code
public static void main(String[] args)
{
int n = 9;
// Function Call
System.out.printf("%d", getNthNumber(n));
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Efficient Python3 program to find n-th palindrome
INT_SIZE = 32
# Construct the nth binary palindrome with the
# given group number, aux_number and operation type
def constructNthNumber(group_no, aux_num, op):
a = [0] * INT_SIZE
num, i = 0, 0
# No need to insert any bit in the middle
if op == 2:
# Length of the final binary representation
len_f = 2 * group_no
# Fill first and last bit as 1
a[len_f - 1] = a[0] = 1
# Start filling the a[] from middle,
# with the aux_num binary representation
while aux_num:
# Get the auxiliary number's ith
# bit and fill around middle
a[group_no + i] = a[group_no - 1 - i] = \
aux_num & 1
aux_num = aux_num >> 1
i += 1
# Insert bit 0 in the middle
elif op == 0:
# Length of the final binary representation
len_f = 2 * group_no + 1
# Fill first and last bit as 1
a[len_f - 1] = a[0] = 1
a[group_no] = 0
# Start filling the a[] from middle, with
# the aux_num binary representation
while aux_num:
# Get the auxiliary number's ith
# bit and fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i] = \
aux_num & 1
aux_num = aux_num >> 1
i += 1
else: # Insert bit 1 in the middle
# Length of the final binary representation
len_f = 2 * group_no + 1
# Fill first and last bit as 1
a[len_f - 1] = a[0] = 1
a[group_no] = 1
# Start filling the a[] from middle, with
# the aux_num binary representation
while aux_num:
# Get the auxiliary number's ith
# bit and fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i] = \
aux_num & 1
aux_num = aux_num >> 1
i += 1
# Convert the number to decimal from binary
for i in range(0, len_f):
num += (1 << i) * a[i]
return num
# Will return the nth binary palindrome number
def getNthNumber(n):
group_no = 0
count_upto_group, count_temp = 0, 1
# Add number of elements in all the groups,
# until the group of the nth number is found
while count_temp < n:
group_no += 1
# Total number of elements until this group
count_upto_group = count_temp
count_temp += 3 * (1 << (group_no - 1))
# Element's offset position in the group
group_offset = n - count_upto_group - 1
# Finding which bit to be placed in the
# middle and finding the number, which we
# will fill from the middle in both directions
if (group_offset + 1) <= (1 << (group_no - 1)):
op = 2 # No need to put extra bit in middle
# We need to fill this auxiliary number
# in binary form the middle in both directions
aux_num = group_offset
else:
if (((group_offset + 1) -
(1 << (group_no - 1))) % 2):
op = 0 # Need to Insert 0 at middle
else:
op = 1 # Need to Insert 1 at middle
aux_num = (((group_offset) -
(1 << (group_no - 1))) // 2)
return constructNthNumber(group_no, aux_num, op)
# Driver code
if __name__ == "__main__":
n = 9
# Function Call
print(getNthNumber(n))
# This code is contributed by Rituraj Jain
C#
// Efficient C# program to find n-th palindrome
using System;
class GFG {
static int INT_SIZE = 32;
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
static int constructNthNumber(int group_no, int aux_num,
int op)
{
int[] a = new int[INT_SIZE];
int num = 0, len_f;
int i = 0;
// No need to insert any bit in the middle
if (op == 2) {
// Length of the final binary representation
len_f = 2 * group_no;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
// Start filling the a[] from middle,
// with the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + i] = a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Insert bit 0 in the middle
else if (op == 0) {
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 0;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
len_f = 2 * group_no + 1;
// Fill first and last bit as 1
a[len_f - 1] = a[0] = 1;
a[group_no] = 1;
// Start filling the a[] from middle, with
// the aux_num binary representation
while (aux_num > 0) {
// Get the auxiliary number's ith bit and
// fill around middle
a[group_no + 1 + i] = a[group_no - 1 - i]
= aux_num & 1;
aux_num = aux_num >> 1;
i++;
}
}
// Convert the number to decimal from binary
for (i = 0; i < len_f; i++)
num += (1 << i) * a[i];
return num;
}
// Will return the nth binary palindrome number
static int getNthNumber(int n)
{
int group_no = 0, group_offset;
int count_upto_group = 0, count_temp = 1;
int op, aux_num;
// Add number of elements in all the groups,
// until the group of the nth number is found
while (count_temp < n) {
group_no++;
// Total number of elements until this group
count_upto_group = count_temp;
count_temp += 3 * (1 << (group_no - 1));
}
// Element's offset position in the group
group_offset = n - count_upto_group - 1;
// Finding which bit to be placed in the
// middle and finding the number, which we
// will fill from the middle in both
// directions
if ((group_offset + 1) <= (1 << (group_no - 1))) {
op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
aux_num = group_offset;
}
else {
if (((group_offset + 1) - (1 << (group_no - 1)))
% 2
== 1)
op = 0; // Need to Insert 0 at middle
else
op = 1; // Need to Insert 1 at middle
aux_num
= ((group_offset) - (1 << (group_no - 1)))
/ 2;
}
return constructNthNumber(group_no, aux_num, op);
}
// Driver code
public static void Main(String[] args)
{
int n = 9;
// Function Call
Console.Write("{0}", getNthNumber(n));
}
}
// This code contributed by Rajput-Ji
的PHP
> 1;
$i++;
}
}
// Insert bit 0 in the middle
else if ($op == 0)
{
// Length of the final binary representation
$len_f = 2 * $group_no + 1;
// Fill first and last bit as 1
$a[$len_f - 1] = $a[0] = 1;
$a[$group_no] = 0;
/* Start filling the a[] from middle, with
the aux_num binary representation */
while ($aux_num)
{
// Get the auxiliary number's ith bit and fill
// around middle
$a[$group_no + 1 + $i] = $a[$group_no - 1 - $i]
= $aux_num & 1;
$aux_num = $aux_num >> 1;
$i++;
}
}
else // Insert bit 1 in the middle
{
// Length of the final binary representation
$len_f = 2 * $group_no + 1;
// Fill first and last bit as 1
$a[$len_f - 1] = $a[0] = 1;
$a[$group_no] = 1;
/* Start filling the a[] from middle, with
the aux_num binary representation */
while ($aux_num)
{
// Get the auxiliary number's ith bit and fill
// around middle
$a[$group_no + 1 + $i] = $a[$group_no - 1 - $i]
= $aux_num & 1;
$aux_num = $aux_num >> 1;
$i++;
}
}
/* Convert the number to decimal from binary */
for ($i = 0; $i < $len_f; $i++)
$num += (1 << $i) * $a[$i];
return $num;
}
/* Will return the nth binary palindrome number */
function getNthNumber($n)
{
$group_no = 0;
$count_upto_group = 0;
$count_temp = 1;
$op=$aux_num=0;
/* Add number of elements in all the groups,
until the group of the nth number is found */
while ($count_temp < $n)
{
$group_no++;
// Total number of elements until this group
$count_upto_group = $count_temp;
$count_temp += 3 * (1 << ($group_no - 1));
}
// Element's offset position in the group
$group_offset = $n - $count_upto_group - 1;
/* Finding which bit to be placed in the
middle and finding the number, which we
will fill from the middle in both
directions */
if (($group_offset + 1) <= (1 << ($group_no - 1)))
{
$op = 2; // No need to put extra bit in middle
// We need to fill this auxiliary number
// in binary form the middle in both directions
$aux_num = $group_offset;
}
else
{
if ((($group_offset + 1) - (1 << ($group_no - 1))) % 2)
$op = 0; // Need to Insert 0 at middle
else
$op = 1; // Need to Insert 1 at middle
$aux_num = (int)((($group_offset) - (1 << ($group_no - 1))) / 2);
}
return constructNthNumber($group_no, $aux_num, $op);
}
// Driver code
$n = 9;
// Function Call
print(getNthNumber($n));
// This code is contributed by mits
?>
输出
27
时间复杂度: O(1)。