lowbit解法
二进制中1的个数
http://www.nowcoder.com/questionTerminal/8ee967e43c2c4ec193b040ea7fbb10b8
class Solution { public: inline int lowBit(int x) { return x & -x; } int NumberOf1(int n) { int cnt = 0; for (; n; ) { n -= lowBit(n); cnt ++; } return cnt; } };