Counting Bits

最新推荐文章于 2025-08-16 00:47:01 发布

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最新推荐文章于 2025-08-16 00:47:01 发布

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分类专栏：算法文章标签： leetcode

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本文介绍了一种使用动态规划解决LeetCode上338题Counting Bits的方法，该方法能在O(n)的时间复杂度内计算出0到给定数num之间所有数的二进制中1的数量。

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leetcode 338. Counting Bits

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.

解题

用一个数组 array 来记录第i个数的二进制中的1数目
首先想到的是枚举时间复杂度为 O(n*sizeof(integer))
然后是递归
当i=0时；array[0]=0
当i=1时；array[1]=1
当i>1时；array[i] = 1 + i - 2^floor(log2(i))

注

floor 是向下取整
log2() 是以2为底求幂

最后递归转换为动态规划，自底向上
时间复杂度为O(n) 空间复杂度为O(n)

代码

public class Solution { public static int[] countBits(int num) { int[] array = new int[num+1]; array[0] = 0; if (num == 0) return array; array[1] = 1; if (num == 1) return array; for (int i=2;i<=num;i++) { int p = i - (int)Math.pow(2,log2floor(i)); array[i] = array[p] + 1; } return array; } // log[b]x = log[a]x / log[a]b static int log2floor(int num){ return (int)(Math.floor(Math.log10(num) / Math.log10(2))); } }