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.

題意，兩個(gè)數(shù)字轉(zhuǎn)為二進(jìn)制，然后異或一下，判斷有多少個(gè)1

直接異或判斷，簡(jiǎn)單粗暴

class Solution {private int ham(int num) {int sum = 0;while (num != 0) {if (num%2 == 1)sum ++;num = num/2;}return sum;}public int hammingDistance(int x, int y) {int res = x^y;return ham(res);} }

?

轉(zhuǎn)載于:https://www.cnblogs.com/Moriarty-cx/p/9769685.html

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