生成具有相同长度的范围 [L, R] 中的所有二进制数
给定两个正整数L和R 。任务是将所有从 L 到 R 的数字转换为二进制数。所有二进制数的长度应该相同。
例子:
Input: L = 2, R = 4
Output:
010
011
100
Explanation: The binary representation of the numbers: 2 = 10, 3 = 11 and 4 = 100.
For the numbers to have same length one preceding 0 is added to the binary representation of both 3 and 4.
Input: L = 2, R = 8
Output:
0010
0011
0100
0101
0110
0111
1000
方法:按照下面提到的方法解决问题。
- 要确定结果二进制数的长度,请取R+1的对数以 2 为底。
- 然后从 L 遍历到 R并将每个数字转换为确定长度的二进制。
- 存储每个数字并在最后打印。
下面是上述方法的实现
C++
// C++ code to implement the approach
#include
using namespace std;
// Function to convert a number to binary
vector convertToBinary(int num,
int length)
{
vector bits(length, 0);
if (num == 0) {
return bits;
}
int i = length - 1;
while (num != 0) {
bits[i--] = (num % 2);
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
vector > getAllBinary(int l,
int r)
{
// Length of the binary numbers
int n = (int) ceil(log(r+1) / log (2));
vector > binary_nos;
for (int i = l; i <= r; i++) {
vector bits =
convertToBinary(i, n);
binary_nos.push_back(bits);
}
return binary_nos;
}
// Driver code
int main()
{
int L = 2, R = 8;
vector > binary_nos =
getAllBinary(L, R);
for (int i = 0; i < binary_nos.size();
i++) {
for (int j = 0; j <
binary_nos[i].size(); j++)
cout << binary_nos[i][j];
cout << endl;
}
return 0;
}
Java
// Java code to implement the approach
import java.util.*;
public class GFG
{
// Function to convert a number to binary
static ArrayList convertToBinary(int num,
int length)
{
ArrayList bits= new ArrayList();
for(int i = 0; i < length; i++) {
bits.add(0);
}
if (num == 0) {
return bits;
}
int i = length - 1;
while (num != 0) {
bits.set(i, (num % 2));
i = i - 1;
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
static ArrayList > getAllBinary(int l,
int r)
{
// Length of the binary numbers
double x = Math.log(r+1);
double y = Math.log (2);
int n = (int) Math.ceil(x / y);
ArrayList > binary_nos =
new ArrayList >();
for (int i = l; i <= r; i++) {
ArrayList bits =
convertToBinary(i, n);
binary_nos.add(bits);
}
return binary_nos;
}
// Driver code
public static void main(String args[])
{
int L = 2, R = 8;
ArrayList > binary_nos =
getAllBinary(L, R);
for (int i = 0; i < binary_nos.size(); i++) {
for (int j = 0; j < binary_nos.get(i).size(); j++) {
System.out.print(binary_nos.get(i).get(j));
}
System.out.println();
}
}
}
// This code is contributed by Samim Hossain Mondal.
Python3
# Python 3 code to implement the approach
import math
# Function to convert a number to binary
def convertToBinary(num, length):
bits = [0]*(length)
if (num == 0):
return bits
i = length - 1
while (num != 0):
bits[i] = (num % 2)
i -= 1
# Integer division
# gives quotient
num = num // 2
return bits
# Function to convert all numbers
# in range [L, R] to binary of
# same length
def getAllBinary(l, r):
# Length of the binary numbers
n = int(math.ceil(math.log(r+1)/ math.log(2)))
binary_nos = []
for i in range(l, r + 1):
bits = convertToBinary(i, n)
binary_nos.append(bits)
return binary_nos
# Driver code
if __name__ == "__main__":
L = 2
R = 8
binary_nos = getAllBinary(L, R)
for i in range(len(binary_nos)):
for j in range(len(binary_nos[i])):
print(binary_nos[i][j], end="")
print()
# This code is contributed by ukasp.
C#
// C# code to implement the approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to convert a number to binary
static List convertToBinary(int num, int length)
{
List bits = new List();
int i;
for (i = 0; i < length; i++)
{
bits.Add(0);
}
if (num == 0)
{
return bits;
}
i = length - 1;
while (num != 0)
{
bits[i] = (num % 2);
i = i - 1;
// Integer division
// gives quotient
num = num / 2;
}
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary of
// same length
static List> getAllBinary(int l, int r)
{
// Length of the binary numbers
double x = Math.Log(r + 1);
double y = Math.Log(2);
int n = (int)Math.Ceiling(x / y);
List> binary_nos = new List>();
for (int i = l; i <= r; i++)
{
List bits = convertToBinary(i, n);
binary_nos.Add(bits);
}
return binary_nos;
}
// Driver code
public static void Main()
{
int L = 2, R = 8;
List> binary_nos = getAllBinary(L, R);
for (int i = 0; i < binary_nos.Count; i++)
{
for (int j = 0; j < binary_nos[i].Count; j++)
{
Console.Write(binary_nos[i][j]);
}
Console.WriteLine("");
}
}
}
// This code is contributed by Saurabh Jaiswal
Javascript
输出
0010
0011
0100
0101
0110
0111
1000
时间复杂度: O(N * logR) 其中 N = (R – L + 1)
辅助空间: O(N * logR)