54.Counting Bits（ 计算1的个数）

Level：

??Medium

题目描述：

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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

思路分析：

??我们可以用动态规划的思想来解决，dp[ i ]，表示，i的二进制中1的个数，那么状态转移方程为dp[ i ]=dp[i>>1]+(i%2)。

代码：

public class Solution{
    public int []countBits(int num){
        int []dp=new int [num+1];
        for(int i=0;i<=num;i++){
            dp[i]=dp[i/2]+i&1;
        }
        return dp;
    }
}

54.Counting Bits（ 计算1的个数）

原文：https://www.cnblogs.com/yjxyy/p/11089017.html