leetcode解题报告--461.Hamming Distance
2017-02-22 17:30
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leetcode解题报告–461.Hamming Distance
作为一个小尝试而已难度是easy
思路:
对2求余数就是二进制时的最低位,因此比较余数就是比较两个数的二进制最低位
比较完最低位后,除于2,相当于右移一位。
如是继续比较,即可。
代码如下:
class Solution { public: int hammingDistance(int x, int y) { int count = 0; while (x > 0 || y > 0) { if (x % 2 != y % 2) { count++; } x = x / 2; y = y / 2; } return count; } };
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