Leetcode 338Counting Bits

问题描述

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.

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

问题分析：利用二进制的特点，所有数都在2^n的某个轮回中，下一个轮回中的1个个数是从0开始到当前轮回开始的那个数为止一一对应的1的个数加1，即：

0 1 2 3 4 5 6 7 8 9

0 1 1 2 1 2 2 3 1 2

1+0 1+1 1+0 1+1 1+1 1+2 1+0 1+1

代码如下：

    public int[] countBits(int num) {        if(num<0)            return null;        int[]dp=new int[num+1];        dp[0]=0;        if(num==0)            return dp;        dp[1]=1;        int flag=2;        int index=0;        for(int i=2;i<=num;i++){            if(index==flag){                flag=flag*2;                index=0;            }            dp[i]=1+dp[index++];        }        return dp;    }