338. Counting Bits--位运算

338. 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.

大致题意： 给定一个数，返回一个数组。这个数组为从零到这个数逐个数字中的二进制数中“1”的个数的排列。

简单的说： 输入‘1’， 返回‘[0,1]’,输入‘2’，返回'[0,1,1]’.就是这样：

public class Solution {    public int[] countBits(int num) {       String s=null;       int count;   int[] result = new int[num+1];       for(int i = 0; i < num+1; i++){      count=0;      s=Integer.toBinaryString(i);      for(int j= 0; j<s.length();j++){      if(s.charAt(j)=='1'){      count++;      }      }      result[i]=count;       }       return result;    }}

使用的是Integer中的二进制转换方法，当然如果使用位运算更好了。

原题链接:https://leetcode.com/problems/counting-bits/