leetcode 338. Counting Bits

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].  

Follow up:

It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
Space complexity should be O(n).
Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.  

思路：dp思想，dp[i] = dp[i&(i-1)] + 1解释一下为什么，dp[i]表示i的二进制1的个数，首先i&(i-1)表示i去掉其二进制最右边的1得到的数x。那么dp[i]肯定是dp[x+1]得到。比如，dp[8] = dp[0] + 1

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ans(num+1, 0);
        for (int i = 1; i <= num; ++i)
            ans[i] = ans[i&(i-1)] + 1;
        return ans;
    }
};