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，這個(gè)效率還不夠高

/*** 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; }

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