Leetcode 191. Number of 1 Bits

题目：

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.

即求出无符号32位整数的二进制数1的位数

最简单的思路既是：右移一位数与1做与运算，结果自加，直至右移32位结束

class Solution {
public:
int hammingWeight(uint32_t n) {
int result = 0, left = 0;
while(n != 0) {
left = n & 0x01;
result += left;
n >>= 1;
}
return result;
}
};

另一种奇妙的解法：参考此博文

n = 0x110100 n-1 = 0x110011 n&(n - 1) = 0x110000

n = 0x110000 n-1 = 0x101111 n&(n - 1) = 0x100000

n = 0x100000 n-1 = 0x011111 n&(n - 1) = 0x0

看到这里已经得到了一种新的解法，n中本来有3个1，按照此种思路只需要循环3此即可求出最终结果

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