LeetCode -- Counting Bits

Question:

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.

Analysis:

给定一个非负的整数num，对于0 ≤ i ≤ num的每一个i，计算i的二进制位中有多少个1，返回一个记录1的个数的数组。

Follow up:

你能在O(n)的时间内只遍历一遍得到答案嘛？

空间复杂度为O(n).

思路：

因为是计算1的个数，所以肯定跟除2以及除2后的余数有关系了。而且计算时是递增的，因此下一个数字可以利用上一个数字的计算结果。

代码如下：

public class Solution {
    public int[] countBits(int num) {
        int[] dp = new int[num+1];
        dp[0] = 0;
        for(int i=1; i<=num; i++) {
            int t1 = i / 2;
            int t2 = i % 2;
            dp[i] = dp[t1] + t2;
        }
        return dp;
    }
}

LeetCode -- Counting Bits

原文：http://www.cnblogs.com/little-YTMM/p/5377489.html