1.题目要求:

两个整数之间的 汉明距离 指的是这两个数字对应二进制位不同的位置的数目。给你两个整数 x 和 y，计算并返回它们之间的汉明距离。示例 1：输入：x = 1, y = 4
输出：2
解释：
1   (0 0 0 1)
4   (0 1 0 0)↑   ↑
上面的箭头指出了对应二进制位不同的位置。
示例 2：输入：x = 3, y = 1
输出：1提示：0 <= x, y <= 231 - 1

2.题目代码:

int hammingDistance(int x, int y) {int x_1 = x;int count_x = 0;//求出x的二进制位的个数while(x_1){count_x++;x_1 /= 2;}int y_1 = y;//求出y的二进制位的个数int count_y = 0;while(y_1){count_y++;y_1 /= 2;}//判断那个个数多if(count_x > count_y){int* number_x =  (int*)malloc(sizeof(int) * count_x);//求出的x的二进制数组int* number_y =  (int*)malloc(sizeof(int) * count_x);//求出的y的二进制数组int j_x = count_x - 1;int j_y = count_x - 1;//把x,y的二进制都存入数组中while(j_x >= 0){number_x[j_x] = x % 2;j_x--;x /= 2;}while(j_y >= 0){number_y[j_y] = y % 2;j_y--;y /= 2;}int diff = 0;//设置变量，找不同for(int i = 0;i < count_x;i++){if(number_x[i] != number_y[i]){diff++;}}return diff;}else{int* number_x = (int*)malloc(sizeof(int) * count_y);//求出的x的二进制数组int* number_y = (int*)malloc(sizeof(int) * count_y);//求出的y的二进制数组int j_x = count_y - 1;int j_y = count_y - 1;//把x,y的二进制都存入数组中while(j_x >= 0){number_x[j_x] = x % 2;j_x--;x /= 2;}while(j_y >= 0){number_y[j_y] = y % 2;j_y--;y /= 2;}int diff = 0;//设置变量，找不同for(int i = 0;i < count_y;i++){if(number_x[i] != number_y[i]){diff++;}}return diff;}
}