leetcode -- 338. Counting Bits 【动态规划】

题目

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.

题意理解

给定一个非负整数 num . 对于每个整数 i 都满足 0 ≤ i ≤ num，计算没个 i 的二进制表示中1 出现的次数，将其作为数组返回。

分析及代码

方法1（思想丰富）：

【机理】将一个十进制转化为2进制，是可以通过除以2来实现的。在这个过程中，i=4,i=2就会有一部分交叉的地方，所以作者从小数算起，为后面的计算提供方便。（减少重复元顺。） 【动态规划】从小数开始算起，后面的结果依赖于前面的结果，将相交叉的部分存储起来。

An easy recurrence for this problem is f[i] = f[i / 2] + i % 2.

public int[] countBits(int num) { int[] f = new int[num + 1]; for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1); return f; }

方法2（常规）：

public int[] countBits(int num) { int[] results = new int[num+1]; for(int i =0;i<num+1;i++){ results[i] = Integer.bitCount(i); } return results; }