letecode [191] - Number of 1 Bits-优快云博客

本文介绍了一种高效算法，用于计算给定二进制整数中1的位数。通过遍历每一位并使用位运算，该算法在C++中实现了快速计算，实测效率高。

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

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

题目大意：

　　给定一个二进制整数，判断其中1 的位数

理解：

　　遍历判断每一位数字是否为1. n&1==1，累加；更新n = n>>1；至n = 0。

代码 C++：

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