位运算

二进制中1的个数

题目链接

每位遍历O(n)

class Solution {
public:
     int  NumberOf1(int n) {
         unsigned int flag=1,sum=0;
         for(int i=0;i<32;++i){
             sum+=flag&n?1:0;
             flag<<=1;
         }
         return sum;
     }
};

只遍历1的个数O(m)

class Solution {
public:
     int  NumberOf1(int n) {
         unsigned int sum=0;
         while(n!=0){
             n=n&(n-1);
             sum++;
         }
         return sum;
     }
};