【动态规划】Leetcode编程题解：338. Counting B 4000 its

题目：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.

样例：For

num = 5

you should return

[0,1,1,2,1,2]

.

要求：

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.

观察每个数字的二进制表达，可以看到每个数字的二进制1的个数都是自己向左挪一位后所得到的数字1的个数加上自己本身最低位构成
代码：

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