检查是否可以用交替的 0 和 1 重新排列二进制字符串

给定一个长度为二进制的字符串，至少两个。我们需要检查是否可以重新排列二进制字符串，使 0 和 1 交替出现。如果可能，则输出为 YES，否则输出为 NO。

例子：

Input : 1011 
Output : NO 
We can’t rearrange the string such that it has alternate 0s and 1s.

Input : 1100 
Output : YES 
There are exactly two ways to rearrange the string, and they are 0101 or 1010 . 

我们可以将所有 0 放在偶数位置，将所有 1 放在奇数位置，或者我们可以将所有 0 放在奇数位置，将所有 1 放在偶数位置。如果字符串的长度是偶数，那么为了满足给定的条件，1 和 0 的计数必须相等。如果字符串的长度是奇数，那么为了满足给定条件，计数的绝对差
1 和 0 必须为 1。

下面是上述方法的实现：

C++

// CPP program to check if we can rearrange a
// string such that it has alternate 0s and 1s.
#include 
using namespace std;
 
// function to check the binary string
bool is_possible(string s)
{
    // length of string
    int l = s.length();
 
    int one = 0, zero = 0;
 
    for (int i = 0; i < l; i++) {
 
        // count zero's
        if (s[i] == '0')
            zero++;
 
        // count one's
        else
            one++;
    }
 
    // if length is even
    if (l % 2 == 0)
        return (one == zero);
 
    // if length is odd
    else
        return (abs(one - zero) == 1);
}
 
// Driver code
int main()
{
    string s = "100110";
    if (is_possible(s))
      cout << "Yes";
    else
      cout << "No";
    return 0;
}

Java

// Java program to check if we can rearrange a
// string such that it has alternate 0s and 1s.
import java.lang.Math;
 
public class GfG{
     
    // function to check the binary string
    public static boolean is_possible(String s)
    {
        // length of string
        int l = s.length();
       
        int one = 0, zero = 0;
       
        for (int i = 0; i < l; i++) {
       
            // count zero's
            if (s.charAt(i) == '0')
                zero++;
       
            // count one's
            else
                one++;
        }
       
        // if length is even
        if (l % 2 == 0) 
            return (one == zero);
       
        // if length is odd
        else
            return (Math.abs(one - zero) == 1);
    }
 
    public static void main(String []args){
         
        String s = "100110";
        if (is_possible(s))
          System.out.println("Yes");
        else
          System.out.println("No");
    }
}
 
// This code is contributed by Rituraj Jain

Python 3

# Python3 program to check if
# we can rearrange a
# string such that it has alternate
# 0s and 1s.
 
# function to check the binary string
def is_possible(s):
 
    # length of string
    l = len(s)
 
    one = 0
    zero = 0
 
    for i in range(0,l) :
 
        # count zero's
        if (s[i] == '0'):
            zero += 1
 
        # count one's
        else:
            one += 1
 
    # if length is even
    if (l % 2 == 0) :
        return (one == zero)
 
    # if length is odd
    else:
        return (abs(one - zero) == 1)
 
# Driver code
if __name__ == "__main__":
    s = "100110"
    if (is_possible(s)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# ChitraNayal

C#

// C# program to check if we can rearrange a
// string such that it has alternate 0s and 1s.
using System;
 
class GfG
{
     
    // function to check the binary string
    public static bool is_possible(String s)
    {
        // length of string
        int l = s.Length;
         
        int one = 0, zero = 0;
         
        for (int i = 0; i < l; i++)
        {
         
            // count zero's
            if (s[i] == '0')
                zero++;
         
            // count one's
            else
                one++;
        }
         
        // if length is even
        if (l % 2 == 0)
            return (one == zero);
         
        // if length is odd
        else
            return (Math.Abs(one - zero) == 1);
    }
 
    // Driver code
    public static void Main(String []args)
    {
         
        String s = "100110";
        if (is_possible(s))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript program to check if we can rearrange a
// string such that it has alternate 0s and 1s.
     
    // function to check the binary string
    function is_possible(s)
    {
        // length of string
        let l = s.length;
         
        let one = 0, zero = 0;
         
        for (let i = 0; i < l; i++) {
         
            // count zero's
            if (s[i] == '0')
                zero++;
         
            // count one's
            else
                one++;
        }
         
        // if length is even
        if (l % 2 == 0) 
            return (one == zero);
         
        // if length is odd
        else
            return (Math.abs(one - zero) == 1);
    }
     
    let s = "100110";
        if (is_possible(s))
          document.write("Yes");
        else
          document.write("No");
       
 
 
// This code is contributed by avanitrachhadiya2155

输出：

Yes

时间复杂度： O(l)，其中 l 是二进制字符串的长度。