【LeetCode & 剑指offer刷题】位运算题2：15 二进制中1的个数（191. Number of 1 Bits）

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191. Number of 1 Bits

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

Example 1:

Input: 11

Output: 3

Explanation: Integer 11 has binary representation 00000000000000000000000000001011

Example 2:

Input: 128

Output: 1

Explanation: Integer 128 has binary representation 00000000000000000000000010000000

 

 

//解法一：利用减一再做与运算可以使最右边的1变为0，统计1的个数

class Solution

{

public:

    int hammingWeight(uint32_t n)

    {

        int count = 0;

        while(n) //有几个1就运算几次，比方法二效率高

        {

            count++;

            n = (n-1) & n; //通过减一再做与运算可以使最右边的1变为0，不断重复可以统计1的个数

        } //最后n变为0

        return count;

    }

};

//解法二：依次和1、10、100...10000000(31个0)做与运算

class Solution

{

public:

    int hammingWeight(uint32_t n)

    {

        int count = 0;

        unsigned int mask = 1; //32位无符号数

        while(mask)

        {

            if(n & mask) count++;

            mask = mask << 1; //左移一位

        }

       

        return count;

    }

};