LeetCode(461) Hamming Distance
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题目
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.
分析
题目含义是求两个整数对应的二进制串中不同比特位值的个数。
可以转化为 求 (x^y)对应二进制串中1的个数。
代码
class Solution { public int hammingDistance(int x, int y) { return countNumOf1Bits(x ^ y); } public int countNumOf1Bits(int n) { int count = 0; while (n > 0) { if ((n & 1) == 1) { ++count; } n >>= 1; } return count; }}
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