leetcode 338. Counting Bits，剑指offer二进制中1的个数

leetcode是求当前所有数的二进制中1的个数，剑指offer上是求某一个数二进制中1的个数

https://www.cnblogs.com/grandyang/p/5294255.html 第三种方法，利用奇偶性找规律

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> result{};
        for(int i = ;i <= num;i++){
            if(i %  == )
                result.push_back(result[i/]);
            else
                result.push_back(result[i/] + );
        }
        return result;
    }
};

class Solution {
public:
     int  NumberOf1(int n) {
         int count = ;
         while(n){
             n = (n-) & n;
             count++;
         }
         return count;
     }
};