leetcode 338. Counting Bits

题目：

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.

   1 public class Solution {
   2     public int[] countBits(int num) {
   3         if(num<0)
   4         {
   5             return null;
   6         }
   7         int[] binaryRepresentation=new int[num+1];
   8         for(int i=0;i<=num;i++)
   9         {
  10             int curNum=i;
  11             while(i>0)
  12             {
  13                 if((i&1)!=0)
  14                 {
  15                     binaryRepresentation[curNum]++;
  16                 }
  17                 i=i>>1;
  18             }
  19             i=curNum;
  20         }
  21         return binaryRepresentation; 
  22     }
  23 }

   

leetcode 338. Counting Bits

原文：http://www.cnblogs.com/HelloWorld-5/p/5413525.html