【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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).
Example 1:
Input: 11
Output: 3
Explanation: Integer 11 has binary representation 00000000000000000000000000001011
Example 2:
Input: 128
Output: 1
Explanation: Integer 128 has binary representation 00000000000000000000000010000000
//解法一：利用减一再做与运算可以使最右边的1变为0，统计1的个数
class Solution
{
public:
int hammingWeight(uint32_t n)
{
int count = 0;
while(n) //有几个1就运算几次，比方法二效率高
{
count++;
n = (n-1) & n; //通过减一再做与运算可以使最右边的1变为0，不断重复可以统计1的个数
} //最后n变为0
return count;
}
};
//解法二：依次和1、10、100...10000000(31个0)做与运算
class Solution
{
public:
int hammingWeight(uint32_t n)
{
int count = 0;
unsigned int mask = 1; //32位无符号数
while(mask)
{
if(n & mask) count++;
mask = mask << 1; //左移一位
}
return count;
}
};