LeetCode 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).

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.

 

分析题目要求计算数字的二进制中1出现的次数，代码如下：

 

 

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

另一种解法，查表法，一个32位整数中有多少个1可以看成4个字节中1的个数相加，结果如下：

unsigned char numof1s[]= {
	0,	1,	1,	2,	1,	2,	2,	3,	1,	2,	2,	3,	2,	3,	3,	4,	
	1,	2,	2,	3,	2,	3,	3,	4,	2,	3,	3,	4,	3,	4,	4,	5,	
	1,	2,	2,	3,	2,	3,	3,	4,	2,	3,	3,	4,	3,	4,	4,	5,	
	2,	3,	3,	4,	3,	4,	4,	5,	3,	4,	4,	5,	4,	5,	5,	6,	
	1,	2,	2,	3,	2,	3,	3,	4,	2,	3,	3,	4,	3,	4,	4,	5,	
	2,	3,	3,	4,	3,	4,	4,	5,	3,	4,	4,	5,	4,	5,	5,	6,	
	2,	3,	3,	4,	3,	4,	4,	5,	3,	4,	4,	5,	4,	5,	5,	6,	
	3,	4,	4,	5,	4,	5,	5,	6,	4,	5,	5,	6,	5,	6,	6,	7,	
	1,	2,	2,	3,	2,	3,	3,	4,	2,	3,	3,	4,	3,	4,	4,	5,	
	2,	3,	3,	4,	3,	4,	4,	5,	3,	4,	4,	5,	4,	5,	5,	6,	
	2,	3,	3,	4,	3,	4,	4,	5,	3,	4,	4,	5,	4,	5,	5,	6,	
	3,	4,	4,	5,	4,	5,	5,	6,	4,	5,	5,	6,	5,	6,	6,	7,	
	2,	3,	3,	4,	3,	4,	4,	5,	3,	4,	4,	5,	4,	5,	5,	6,	
	3,	4,	4,	5,	4,	5,	5,	6,	4,	5,	5,	6,	5,	6,	6,	7,	
	3,	4,	4,	5,	4,	5,	5,	6,	4,	5,	5,	6,	5,	6,	6,	7,	
	4,	5,	5,	6,	5,	6,	6,	7,	5,	6,	6,	7,	6,	7,	7,	8,	
};

int hammingWeight(uint32_t n) {
       return numof1s[n&0xff] + numof1s[(n>>8)&0xff] +
		numof1s[(n>>16)&0xff] + numof1s[(n>>24)];
    }