leetcode338—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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

想法：统计0-num之间的二进制表示中1的个数。用result[i]表示i的二进制表示中1的个数。利用动态规划，找出状态转移表达式result[i]=result[i&i-1]+1

class Solution {
public:
    vector<int> countBits(int num) {
        vector<);
        result[] = ;
         ;i <= num ; i++){
            result[i] = result[i & i-] + ;
        }
        return result;
    }
};