Counting Bits问题及解法

问题描述：

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.

示例：
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.

问题分析：

我们知道，计算一个数n的二进制表示中‘1’的个数时，我们利用 n = n &(n - 1)进行梯度下降求解，经过分析可知，dp[i] = dp[i&(i - 1)] + 1.

过程详见代码：

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