生成所有长度为 N 的二进制字符串，0 和 1 的计数相等
原文:https://www . geesforgeks . org/generate-all-binary-length-n-with-count-0s-1s/
给定一个整数 N ，任务是生成所有二进制字符串，其中等于T6】0s 和 1s 。如果没有字符串，打印 -1
示例:
输入 : N = 2 输出:【01】【10】 解释:长度为 2 的所有可能的二进制字符串为:01、10、11、00。其中，只有 2 个具有相同数量的 0 和 1
输入: 4 输出:“0011”“0101”“0110”“1100”“1010”“1001”
方法:使用递归可以解决任务。如果 N 为奇数，那么答案为 -1 ，否则，我们可以使用递归生成所有等于0 和 1 的二进制字符串。按照以下步骤解决问题:
- 变量1记录字符串中1的数量，变量0记录字符串中0的数量。
- 1和0都应该有频率 N/2 。
- 基础条件:字符串 s 存储输出字符串。因此，当 s 的长度达到 N 时，我们停止递归调用并打印输出字符串 s 。
- 如果 1 的的频率小于的 N/2 ，则将 1 加到弦上，递增 1。
- 如果 0 的的频率小于的 N/2 ，则在字符串中添加 0 ，并且递增零。
下面是上述代码的实现:
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Recursive function that prints
// all strings of N length with equal 1's and 0's
void binaryNum(int n, string s, int ones,
int zeros)
{
// String s contains the output to be printed
// ones stores the frequency of 1's
// zeros stores the frequency of 0's
// Base Condition: When the length of string s
// becomes N
if (s.length() == n)
{
cout << (s) << endl;
return;
}
// If frequency of 1's is less than N/2 then
// add 1 to the string and increment ones
if (ones < n / 2)
binaryNum(n, s + "1", ones + 1, zeros);
// If frequency of 0's is less than N/2 then
// add 0 to the string and increment zeros
if (zeros < n / 2)
binaryNum(n, s + "0", ones, zeros + 1);
}
// Driver Code
int main()
{
string s = "";
binaryNum(4, s, 0, 0);
return 0;
}
// This code is contributed by Potta Lokesh
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG {
// Recursive function that prints
// all strings of N length with equal 1's and 0's
static void binaryNum(int n, String s, int ones,
int zeros)
{
// String s contains the output to be printed
// ones stores the frequency of 1's
// zeros stores the frequency of 0's
// Base Condition: When the length of string s
// becomes N
if (s.length() == n) {
System.out.println(s);
return;
}
// If frequency of 1's is less than N/2 then
// add 1 to the string and increment ones
if (ones < n / 2)
binaryNum(n, s + "1", ones + 1, zeros);
// If frequency of 0's is less than N/2 then
// add 0 to the string and increment zeros
if (zeros < n / 2)
binaryNum(n, s + "0", ones, zeros + 1);
}
// Driver Code
public static void main(String[] args)
{
String s = "";
binaryNum(4, s, 0, 0);
}
}
Python 3
# python code for the above approach
# Recursive function that prints
# all strings of N length with equal 1's and 0's
def binaryNum(n, s, ones, zeros):
# String s contains the output to be printed
# ones stores the frequency of 1's
# zeros stores the frequency of 0's
# Base Condition: When the length of string s
# becomes N
if (len(s) == n):
print(s)
return
# If frequency of 1's is less than N/2 then
# add 1 to the string and increment ones
if (ones < n / 2):
binaryNum(n, s + "1", ones + 1, zeros)
# If frequency of 0's is less than N/2 then
# add 0 to the string and increment zeros
if (zeros < n / 2):
binaryNum(n, s + "0", ones, zeros + 1)
# Driver Code
if __name__ == "__main__":
s = ""
binaryNum(4, s, 0, 0)
# This code is contributed by rakeshsahni
C
// C# program for the above approach
using System;
class GFG {
// Recursive function that prints
// all strings of N length with equal 1's and 0's
static void binaryNum(int n, string s, int ones,
int zeros)
{
// String s contains the output to be printed
// ones stores the frequency of 1's
// zeros stores the frequency of 0's
// Base Condition: When the length of string s
// becomes N
if (s.Length == n) {
Console.WriteLine(s);
return;
}
// If frequency of 1's is less than N/2 then
// add 1 to the string and increment ones
if (ones < n / 2)
binaryNum(n, s + "1", ones + 1, zeros);
// If frequency of 0's is less than N/2 then
// add 0 to the string and increment zeros
if (zeros < n / 2)
binaryNum(n, s + "0", ones, zeros + 1);
}
// Driver Code
public static void Main(string[] args)
{
string s = "";
binaryNum(4, s, 0, 0);
}
}
// This code is contributed by ukasp.
java 描述语言
<script>
// javascript program for the above approach
// Recursive function that prints
// all strings of N length with equal 1's and 0's
function binaryNum(n, s, ones, zeros)
{
// String s contains the output to be printed
// ones stores the frequency of 1's
// zeros stores the frequency of 0's
// Base Condition: When the length of string s
// becomes N
if (s.length == n) {
document.write(s+"<br>");
return;
}
// If frequency of 1's is less than N/2 then
// add 1 to the string and increment ones
if (ones < n / 2)
binaryNum(n, s + "1", ones + 1, zeros);
// If frequency of 0's is less than N/2 then
// add 0 to the string and increment zeros
if (zeros < n / 2)
binaryNum(n, s + "0", ones, zeros + 1);
}
// Driver Code
var s = "";
binaryNum(4, s, 0, 0);
// This code is contributed by 29AjayKumar
</script>
Output
1100
1010
1001
0110
0101
0011
时间复杂度:O(2N) 辅助空间 : O(1)
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