# 重新排列二进制字符串作为 x 和 y 的交替出现 > 原文： [https://www.geeksforgeeks.org/rearrange-binary-string-alternate-x-y-occurrences/](https://www.geeksforgeeks.org/rearrange-binary-string-alternate-x-y-occurrences/) 给定一个二进制字符串`s`，并给出两个整数`x`和`y`。 任务是按照给定的字符串进行排列，以使 0 代表`X`，然后 1 代表`Y`，依此类推，直到完成 0 或 1 之一。 然后将字符串的其余部分连接起来并打印最终的字符串。 给出：`x`或`y`不能为 0 例子： ``` Input : s = "0011" x = 1 y = 1 Output : 0101 x is 1 and y is 1\. So first we print '0' one time the '1' one time and then we print '0', after printing '0', all 0's are vanished from the given string so we concatenate rest of the string which is '1'. Input : s = '1011011' x = 1 y = 1 Output : 0101111 ``` 1. 计算字符串中 0 和 1 的数量。 2. 循环运行，直到两个字母之一完成。 1. 首先打印 0 直到`x`且递减计数为 0。 2. 然后打印 1 直到`y`且递减计数为 0。 ## C++ ```cpp // C++ program to arrange given string #include <bits/stdc++.h> using namespace std; // Function which arrange the given string void arrangeString(string str, int x, int y) {     int count_0 = 0;     int count_1 = 0;     int len = str.length();     // Counting number of 0's and 1's in the     // given string.     for (int i = 0; i < len; i++) {         if (str[i] == '0')             count_0++;         else             count_1++;     }     // Printing first all 0's x-times     // and decrement count of 0's x-times     // and do the similar task with '1'     while (count_0 > 0 || count_1 > 0) {      for (int j = 0; j < x && count_0 > 0; j++) {         if (count_0 > 0) {            cout << "0";            count_0--;         }      }      for (int j = 0; j < y && count_1 > 0; j++) {         if (count_1 > 0) {            cout << "1";            count_1--;         }       }     } } // Driver Code int main() {     string str = "01101101101101101000000";     int x = 1;     int y = 2;     arrangeString(str, x, y);     return 0; } ```