Leetcode 191 Number of 1 Bits

triplebee

发布于 2018-01-12 14:46:32

5340

发布于 2018-01-12 14:46:32

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.

统计bit中1的数量。

简单做法，我这样每一位都搞一遍。

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

优化做法，摘选自维基百科，注释说的很清楚。

有时候觉得这类题目挺有意思的，好像被小学奥数题难倒的感觉

有时候又觉得这类题目没有意思，好像计算机学生研究哲学一样？哈哈哈！

// This is a naive implementation, shown for comparison, and to help in understanding the better functions. 
// It uses 24 arithmetic operations (shift, add, and).
int hammingWeight(uint32_t n)
{
    n = (n & 0x55555555) + (n >>  1 & 0x55555555); // put count of each  2 bits into those  2 bits 
    n = (n & 0x33333333) + (n >>  2 & 0x33333333); // put count of each  4 bits into those  4 bits 
    n = (n & 0x0F0F0F0F) + (n >>  4 & 0x0F0F0F0F); // put count of each  8 bits into those  8 bits 
    n = (n & 0x00FF00FF) + (n >>  8 & 0x00FF00FF); // put count of each 16 bits into those 16 bits 
    n = (n & 0x0000FFFF) + (n >> 16 & 0x0000FFFF); // put count of each 32 bits into those 32 bits 
    return n;
}

// This uses fewer arithmetic operations than any other known implementation on machines with slow multiplication.
// It uses 17 arithmetic operations.
int hammingWeight(uint32_t n)
{
    n -= (n >> 1) & 0x55555555; //put count of each 2 bits into those 2 bits
    n = (n & 0x33333333) + (n >> 2 & 0x33333333); //put count of each 4 bits into those 4 bits
    n = (n + (n >> 4)) & 0x0F0F0F0F; //put count of each 8 bits into those 8 bits
    n += n >> 8; // put count of each 16 bits into those 8 bits
    n += n >> 16; // put count of each 32 bits into those 8 bits
    return n & 0xFF;
}

// This uses fewer arithmetic operations than any other known implementation on machines with fast multiplication.
// It uses 12 arithmetic operations, one of which is a multiply.
int hammingWeight(uint32_t n)
{
    n -= (n >> 1) & 0x55555555; // put count of each 2 bits into those 2 bits
    n = (n & 0x33333333) + (n >> 2 & 0x33333333); // put count of each 4 bits into those 4 bits
    n = (n + (n >> 4)) & 0x0F0F0F0F; // put count of each 8 bits into those 8 bits 
    return n * 0x01010101 >> 24; // returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24)
}

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原始发表：2017-03-12 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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