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：

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

Note：

• 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.

解釋下題目：

給定一個(gè)正數(shù)，然后返回它之前每一個(gè)數(shù)字中（二進(jìn)制）的1的個(gè)數(shù)

1. 動(dòng)態(tài)規(guī)劃

實(shí)際耗時(shí)：1ms

public int[] countBits(int num) {
    int[] res = new int[num + 1];
    for (int i = 1; i <= num; i++) {
        res[i] = res[i >> 1] + (i & 1);
    }
    return res;
}

??那這種題目肯定是動(dòng)態(tài)規(guī)劃，重要的是理解的是一個(gè)數(shù)n的兩倍其實(shí)就是后面加個(gè)0，所以就有了這個(gè)算法

時(shí)間復(fù)雜度O(n)

空間復(fù)雜度O(n)