题目描述

• 涉及二进制，位运算跑不了～
  

思路 & 代码

• 既然是32位，那么通过一次遍历，每次判断一个位是否为1即可

public class Solution {// you need to treat n as an unsigned valuepublic int hammingWeight(int n) {int ans = 0;// 遍历各个位，进行“截取”操作for(int i = 0; i < 32; i++){// “截取”到了，说明这里有1if(((1 << i) & n) != 0){ans++;}}return ans;}
}

二刷

• 耗时稍微快了一点呢！

public class Solution {// you need to treat n as an unsigned valuepublic int hammingWeight(int n) {int counts = 0;for(int i = 0; i < 32; i++, n >>= 1) {if((n & 1) == 1) {counts++;}}return counts;}
}