剑指 Offer 15. 二进制中1的个数

image.png
image.png

将使用与运算检查每一位

  1. public class Solution {
  2.     public int hammingWeight(int n) {
  3.         int count = 0;
  4.         for (int i = 0; i < 32; i++) {
  5.             if ((n >> i & 1) != 0) count ++;
  6.         }
  7.         return count;
  8.     }
  9. }

巧妙地用 n 与 n - 1 对比

剑指 Offer 15. 二进制中1的个数 - 图3

图片来源 https://leetcode-cn.com/problems/er-jin-zhi-zhong-1de-ge-shu-lcof/solution/mian-shi-ti-15-er-jin-zhi-zhong-1de-ge-shu-wei-yun