Dynamic Programing -- Leetcode problem 338. Counting Bits

• 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.

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

• 分析：这道题从pass起来非常容易，但是关键在与如何设计出是件复杂度为O(n)的算法。
• 思路一：直接使用循环暴力求解。O(n*sizeof(integer))

class Solution {public:    vector<int> countBits(int num) {        vector<int> my_vec;        for (int i = 0; i <= num; i++) {            my_vec.push_back(count_num(i));        }        return my_vec;    }    int count_num(int num) {        int count = 0;        int find_num = 0;        while (num != 0) {            find_num = num % 2;            if (find_num == 1) {                count++;            }            num = num / 2;        }        return count;    }};

• 思路二：使用动态规划的思想求解。(O(n))

十进制 二进制 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111 … …

对于这些二进制的值，先出现的二进制会在后出现的二进制中被利用，因此我们在计算中只需要将利用前面统计出来的结果加上最低位的1就可以完成。

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