Leetcode: 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: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.

Hint:

You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

1 public class Solution {
2     public int[] countBits(int num) {
3         int[] res = new int[num+1];
4         for (int i=1; i<=num; i++) {
5             res[i] = res[i>>1] + (i&1);
6         }
7         return res;
8     }
9 }

转载于:https://www.cnblogs.com/EdwardLiu/p/6092304.html

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