[LeetCode]338. Counting Bits

Problem Description

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.
[https://leetcode.com/problems/counting-bits/]

思路

DP。
对于一个数，如果他是偶数，那它二进制后含有‘1’的个数与左移一位之后含有‘1’的个数相同，如果是奇数，则加1。
例如：
偶数1100110中含有的1与110011相同
奇数1100111中含有的1比110011多1.

Code

package q338;public class Solution {    public int[] countBits(int num) {        int[] ans=new int[num+1];        ans[0]=0;        for(int i=1;i<=num;i++){            ans[i]=ans[i-1]>>1+i%2;        }        return ans;    }}