Counting Bits

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本文介绍了一种算法，用于计算从0到给定非负整数范围内每个数的二进制表示中1的数量，并返回这些数量组成的数组。讨论了如何优化算法以达到线性时间复杂度，并提供了一个不使用内置函数的实现示例。

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

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].

Follow up:

1、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?
2、Space complexity should be O(n).
3、Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

方案一，这个效率还不够高

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* countBits(int num, int* returnSize) {
     int i;
     unsigned flag;

    *returnSize = num + 1;
    int *a = (int *)malloc((*returnSize) * sizeof(int));
    a[0] = 0;

     flag = 0;
     for(i = 1; i <= num; ++i) {
         flag = i;
         a[i] = 0;
         while(flag > 0) {
             flag = flag & (flag - 1);
             a[i]++;
         }
    }

    return a;
}