[2018-12-16] [LeetCode-Week15] 338. Counting Bits 递推

https://leetcode.com/problems/counting-bits/

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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.

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对于每个数，获得他右移之后得到的数的 1 的个数，然后求奇偶看其最后一位是 1 还是 0 即可。

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class Solution {
public:
    vector countBits(int num) {
        vector d;
        d.push_back(0);
        
        for (int i = 1; i <= num; i++) {
            int nd = d[i>>1] + i%2;
            d.push_back(nd);
        }
        return d;
    }
    
};