題目

Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the?Hamming weight).

For example, the 32-bit integer ’11' has binary representation?00000000000000000000000000001011, so the function should return 3.

Credits:
Special thanks to?@ts?for adding this problem and creating all test cases.

代碼

1.遞歸運算，雖然leet不通過，因為時間原因 public class Solution {// you need to treat n as an unsigned valuepublic static int hammingWeight(int n) {if(n==0) return 0;if(n==1) return 1;int total=1;int index=1;while(n-total*2>=0){total=total*2;}n=n-total;index=index+hammingWeight(n);return index;}

2.用位運算方法

public class Solution {// you need to treat n as an unsigned valuepublic int hammingWeight(int n) {if(n==0) return 0;int i=0;while(n!=0){n=n & (n-1);i++;}return i;} }

代碼下載：https://github.com/jimenbian/GarvinLeetCode

/********************************

* 本文來自博客 ?“李博Garvin“

* 轉載請標明出處:http://blog.csdn.net/buptgshengod

******************************************/

總結

以上是生活随笔為你收集整理的【LeetCode从零单排】No 191.Number of 1 Bits(考察位运算)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。