LeetCode Hamming Distance The Hamming distance 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 = 4
Output: 2
Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑
The above arrows point to positions where the corresponding bits are different.

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求两个整数的汉明距离，即两个整数的二进制串中不一样的位数。马上联想到之前LeetCode Number of 1 Bits求二进制中1的个数，这里是一样的，相当于求两个整数的二进制位，每次比较一下，如果不同，则汉明距离加1；当发现两个数已经相等时，汉明距离不可能再增加了。 完整代码如下： [cpp] class Solution { public: int hammingDistance(int x, int y) { int dist = 0; for (int i = 0; i < 32; i++) { if (x == y)break; if ((x & 1) != (y & 1))dist++; x >>= 1; y >>= 1; } return dist; } }; [/cpp] 本代码提交AC，用时0MS，击败55.12%，加了那个break效率会高很多。]]>