生成所有长度为 N 且 0 和 1 计数相等的二进制字符串

给定一个整数N ，任务是生成所有等于0和1的二进制字符串。如果没有字符串是可能的，打印-1

例子：

Input: N = 2  
Output: “01”, “10”
Explanation: All possible binary strings of length 2 are: 01, 10, 11, 00. Out of these, only 2 have equal number of 0s and 1s  

Input: 4  
Output:  “0011”, “0101”, “0110”, “1100”, “1010”, “1001” 

方法：可以使用递归来解决该任务。如果N为奇数，则答案为-1 ，否则，我们可以使用递归来生成所有等于0 和 1 的二进制字符串。请按照以下步骤解决问题：

• 变量 one 跟踪1 的数量，变量zeros跟踪字符串中0 的数量。
• 一和零都应该有频率N/2 。
• 基本条件：字符串s存储输出字符串。所以，当 s 的长度达到N我们停止递归调用并打印输出字符串s 。
• 如果1 的频率小于N/2 ，则将1加到字符串中并增加 1 。
• 如果0 的频率小于N/2 ，则将0添加到字符串并增加 zeros 。

下面是上述代码的实现：

C++

// C++ code for the above approach
#include 
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

// 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);
    }
}

Python3

# 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.

Javascript

// 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

输出

1100
1010
1001
0110
0101
0011

时间复杂度： O(2 ^N )
辅助空间：O(1)