给定一个整数N ，任务是找到最小的整数X ，使其不设置奇数位并且X≥N 。
注意：假定从右侧开始定位位，并且假定第一位为第0位。
例子：
Input: N = 9
Output: 16
16’s binary representation is 10000, which has its 4th bit
set which is the smallest number possible satisfying the given condition.
Input: N = 5
Output: 5
Input: N = 19
Output: 20
方法：可以使用贪婪方法和一些位属性来解决问题。如果将较小的2的幂恰好取一次并相加，则它们永远不能超过较高的2的幂(例如(1 + 2 + 4)<8)。以下贪婪方法用于解决上述问题：
- 最初计算位数。
- 获取设置位的最左索引。
- 如果最左边的设置位处于奇数索引，则答案将始终为(1 <<(leftmost_bit_index + 1)) 。
- 否则，通过将所有偶数位设置为0到leftmost_bit_index，贪婪地形成一个数字。现在从右边贪婪地删除2的幂以检查我们是否得到满足指定条件的数字。
- 如果上述条件不能提供我们的电话号码，那么我们只需设置下一个最左边的偶数位并返回答案。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to count the total bits
int countBits(int n)
{
int count = 0;
// Iterate and find the
// number of set bits
while (n) {
count++;
// Right shift the number by 1
n >>= 1;
}
return count;
}
// Function to find the nearest number
int findNearestNumber(int n)
{
// Count the total number of bits
int cnt = countBits(n);
// To get the position
cnt -= 1;
// If the last set bit is
// at odd position then
// answer will always be a number
// with the left bit set
if (cnt % 2) {
return 1 << (cnt + 1);
}
else {
int tempnum = 0;
// Set all the even bits which
// are possible
for (int i = 0; i <= cnt; i += 2)
tempnum += 1 << i;
// If the number still is less than N
if (tempnum < n) {
// Return the number by setting the
// next even set bit
return (1 << (cnt + 2));
}
else if (tempnum == n)
return n;
// If we have reached this position
// it means tempsum > n
// hence turn off even bits to get the
// first possible number
for (int i = 0; i <= cnt; i += 2) {
// Turn off the bit
tempnum -= (1 << i);
// If it gets lower than N
// then set it and return that number
if (tempnum < n)
return tempnum += (1 << i);
}
}
}
// Driver code
int main()
{
int n = 19;
cout << findNearestNumber(n);
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to count the total bits
static int countBits(int n)
{
int count = 0;
// Iterate and find the
// number of set bits
while (n > 0)
{
count++;
// Right shift the number by 1
n >>= 1;
}
return count;
}
// Function to find the nearest number
static int findNearestNumber(int n)
{
// Count the total number of bits
int cnt = countBits(n);
// To get the position
cnt -= 1;
// If the last set bit is
// at odd position then
// answer will always be a number
// with the left bit set
if (cnt % 2 == 1)
{
return 1 << (cnt + 1);
}
else
{
int tempnum = 0;
// Set all the even bits which
// are possible
for (int i = 0; i <= cnt; i += 2)
{
tempnum += 1 << i;
}
// If the number still is less than N
if (tempnum < n)
{
// Return the number by setting the
// next even set bit
return (1 << (cnt + 2));
}
else
if (tempnum == n)
{
return n;
}
// If we have reached this position
// it means tempsum > n
// hence turn off even bits to get the
// first possible number
for (int i = 0; i <= cnt; i += 2)
{
// Turn off the bit
tempnum -= (1 << i);
// If it gets lower than N
// then set it and return that number
if (tempnum < n)
{
return tempnum += (1 << i);
}
}
}
return Integer.MIN_VALUE;
}
// Driver code
public static void main(String[] args)
{
int n = 19;
System.out.println(findNearestNumber(n));
}
}
// This code is contributed by 29AjayKumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to count the total bits
static int countBits(int n)
{
int count = 0;
// Iterate and find the
// number of set bits
while (n > 0)
{
count++;
// Right shift the number by 1
n >>= 1;
}
return count;
}
// Function to find the nearest number
static int findNearestNumber(int n)
{
// Count the total number of bits
int cnt = countBits(n);
// To get the position
cnt -= 1;
// If the last set bit is
// at odd position then
// answer will always be a number
// with the left bit set
if (cnt % 2 == 1)
{
return 1 << (cnt + 1);
}
else
{
int tempnum = 0;
// Set all the even bits which
// are possible
for (int i = 0; i <= cnt; i += 2)
{
tempnum += 1 << i;
}
// If the number still is less than N
if (tempnum < n)
{
// Return the number by setting the
// next even set bit
return (1 << (cnt + 2));
}
else
if (tempnum == n)
{
return n;
}
// If we have reached this position
// it means tempsum > n
// hence turn off even bits to get the
// first possible number
for (int i = 0; i <= cnt; i += 2)
{
// Turn off the bit
tempnum -= (1 << i);
// If it gets lower than N
// then set it and return that number
if (tempnum < n)
{
return tempnum += (1 << i);
}
}
}
return int.MinValue;
}
// Driver code
public static void Main()
{
int n = 19;
Console.WriteLine(findNearestNumber(n));
}
}
// This code is contributed by anuj_67..Python3
# Python implementation of the above approach
# Function to count the total bits
def countBits(n):
count = 0;
# Iterate and find the
# number of set bits
while (n>0):
count+=1;
# Right shift the number by 1
n >>= 1;
return count;
# Function to find the nearest number
def findNearestNumber(n):
# Count the total number of bits
cnt = countBits(n);
# To get the position
cnt -= 1;
# If the last set bit is
# at odd position then
# answer will always be a number
# with the left bit set
if (cnt % 2):
return 1 << (cnt + 1);
else:
tempnum = 0;
# Set all the even bits which
# are possible
for i in range(0,cnt+1,2):
tempnum += 1 << i;
# If the number still is less than N
if (tempnum < n):
# Return the number by setting the
# next even set bit
return (1 << (cnt + 2));
elif (tempnum == n):
return n;
# If we have reached this position
# it means tempsum > n
# hence turn off even bits to get the
# first possible number
for i in range(0,cnt+1,2):
# Turn off the bit
tempnum -= (1 << i);
# If it gets lower than N
# then set it and return that number
if (tempnum < n):
tempnum += (1 << i);
return tempnum;
# Driver code
n = 19;
print(findNearestNumber(n));
# This code contributed by PrinciRaj1992输出：
20