338. Counting Bits

思路：动态规划

先找规律：每逢数字是2的次方，均为1

递推方程：dp[i]=dp[i-k]+dp[k];其中k是小于i的最近一个2的整数次方

 

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res;
        int k=2;
        if(num<0) return res;
        
        res.push_back(0);if(num==0) return res;
        res.push_back(1);if(num==1) return res;
        
        for(int i=2;i<=num;i++){
            if(i%k==0){res.push_back(1);k=i;continue;}//开始写错了，写成i%2==0,导致bug
            //if(i==7)cout<<i-k<<" "<<res[i-k]<<" "<<k<<""<<res[k];
            res.push_back(res[i-k]+res[k]);
        }
        return res;
    }
};