[LeetCode] 461.Hamming Distance 备忘
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461. 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 = 4Output: 2Explanation:1 (0 0 0 1)4 (0 1 0 0) ? ?The above arrows point to positions where the corresponding bits are different.
求两个整数之间对应的位上,不相同的位有多少个。
由于题目中对于x,y限制了大小,所以最简单的方法,就是2个数字异或之后做右位移运算,判断每一个位置是否为1,进行计数,直到右移次数到达31次为止。
class Solution {public: int hammingDistance(int x, int y) { int n = x ^ y; int i = 0; int distance = 0; do { if (n & 1 == 1) { ++distance; } n >>= 1; ++i; }while (i < 31); return distance; }};
class Solution {public: int hammingDistance(int x, int y) { int n = x ^ y; int count = 0; while (n) { count++; n = n & n - 1; } return count; }};
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