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 }

?

轉(zhuǎn)載于:https://www.cnblogs.com/EdwardLiu/p/6092304.html

總結(jié)

以上是生活随笔為你收集整理的Leetcode: Counting Bits的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。