【LeetCode-338】 Counting Bits

原题：

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:

• 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.

题目大意：

给一个非负整数num，输出[0,num]间每一个数字的二进制表示中1的个数。

思路：

把前几个值动手写一写，很快就能发现迭代规律，构建出如下图的二叉树。

代码：

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */int* countBits(int num, int* returnSize) {    int* p;    *returnSize = num+1;    p = (int*)malloc(*returnSize*sizeof(int));    if (num == 0){        p[0]=0;        return p;    }    p[0]=0;    p[1]=1;    for (int i = 2 ; i <= num ; i++){        if (i % 2 == 0){            p[i] = p[i/2];        }else{            p[i] = p[i/2]+1;        }    }    return p;}