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) }