【LeetCode从零单排】No 191.Number of 1 Bits(考察位运算)-优快云博客

本文介绍了一种计算32位整数中1的位数的方法，提供了两种实现方案：一种是递归方法，但由于时间复杂度较高无法通过LeetCode测试；另一种是使用位运算，效率较高。

题目

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.

代码

1.递归运算，虽然leet不通过，因为时间原因

public class Solution {
    // you need to treat n as an unsigned value
    public 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.用位运算方法

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

代码下载：https://github.com/jimenbian/GarvinLeetCode

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

* 本文来自博客 “李博Garvin“

* 转载请标明出处:http://blog.youkuaiyun.com/buptgshengod

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