算法复习：位运算

Posted 2021-03-10 dzzy

leetcode 191. 位1的个数

维护匹配串 从1开始逐位移动比较

class Solution {
public:
    int hammingWeight(uint32_t n) 
    {     // uint32_t为32位无符号类型数据
        int count=0;
        uint32_t donser=1;
        for(int i=0;i<32;i++)
        {
            if((n&donser)>=1)
                count++;
            donser<<=1;
        }
        return count;
    }
};

leedcode 191

n与n-1取& 

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

leedcode 191

调用内置函数 __builtin_popcount(n)

class Solution {
public:
    int hammingWeight(uint32_t n) 
    {     // uint32_t为32位无符号类型数据
        return __builtin_popcount(n);
    }
};