Leetcode 191. Number of 1 Bits

191. Number of 1 Bits

• Total Accepted: 103908
• Total Submissions: 275693
• Difficulty: Easy

 

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.

 

思路：

方法一：通过不断和1与操作求出数的二进制中1的个数。
方法二：利用n&(n-1)。

n&(n-1)可以用来做以下件事：
1. 计算数字i的二进制表示中1的个数。

while (n >0 ) {
      count ++;
      n &= (n-1);
}

 

2. 判断i是不是2的整数次幂。

n > 0 && ((n & (n - 1)) == 0 )

 

3. 

 

代码：

方法一：

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count=0;
        while(n){
            count+=n&1;
            n=n>>1;
        }
        return count;
    }
};

 

方法二：

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count=0;
        while(n){
            n&=n-1;
            count++;
        }
        return count;
    }
};