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题目：
The between two integers is the number of positions at which the corresponding bits are different.
Given two integers x and y, calculate the Hamming distance.
Note:
0 ≤ x, y < 231.
Example:
Input: x = 1, y = 4Output: 2Explanation:1 (0 0 0 1)4 (0 1 0 0) ↑ ↑The above arrows point to positions where the corresponding bits are different.思路：
求两个数的汉明距离，用取余方法，算出x的二进制表示结果，若不到32位，则补0。并按位比较x与y
代码：
class Solution {public: int hammingDistance(int x, int y) { vector x1, y1; int count = 0; int count_x = 0; int count_y = 0; while (x>0){//用取余方法，算出x的二进制表示结果 x1.push_back(x % 2); x = x / 2; count_x++; } while (count_x<31){//若不到32位，则补0 x1.push_back(0); count_x++; } while (y>0){//与x同样 y1.push_back(y % 2); y = y / 2; count_y++; } while (count_y < 31){ y1.push_back(0); count_y++; } for (int i = 0; i != 31; i++){ if (x1[i] ^ y1[i]){//按位比较x与y的 count++; } } return count; }}; 转载地址：http://cnmii.baihongyu.com/