Leet Code OJ 338. Counting Bits [Difficulty: Medium]

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.
Hint:
You should make use of what you have produced already.

翻译：
给定一个非负整数num，对于每个0<=i<=num的整数i，计算i的二进制表示中1的个数，返回这些个数作为一个数组。
例如，输入num = 5 你应该返回 [0,1,1,2,1,2].

分析：
按照常规思路，很容易得出“Java代码2”的方案，但是这个方案的时间复杂度是O（nlogn）。
通过对数组的前64个元素进行分析（num=63），我们发现数组呈现一定的规律，不断重复，如下图所示：

0
1
1 2
1 2 2 3
1 2 2 3 2 3 3 4
1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5
1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5 2 3 3 4 3 4 4 5 3 4 4 5 4 5 5 6

由此我们发现0112是一个基础元素，不断循环反复，可以推论：如果已知第一个元素是result[0],那么第二第三个元素为result[0]+1，第四个元素为result[0]+2，由此获得前4个元素result[0]~result[3]；以这4个元素为基础，我们可以得到
result[4]=result[0]+1,result[5]=result[1]+1…，
result[8]=result[0]+1,result[9]=result[1]+1… ,
result[12]=result[0]+2,result[13]=result[1]+2…;
以此类推可以获得全部的数组。

Java版代码1：

public class Solution {public int[] countBits(int num) {int[] result = new int[num + 1];int range = 1;result[0] = 0;boolean stop = false;while (!stop) {stop = fillNum(result, range);range *= 4;}return result;}public boolean fillNum(int[] nums, int range) {for (int i = 0; i < range; i++) {if (range + i < nums.length) {nums[range + i] = nums[i] + 1;} else {return true;}if (2 * range + i < nums.length) {nums[2 * range + i] = nums[i] + 1;}if (3 * range + i < nums.length) {nums[3 * range + i] = nums[i] + 2;}}return false;}
}

Java版代码2：

public class Solution {public int[] countBits(int num) {int[] result=new int[num+1];result[0]=0;for(int i=1;i<=num;i++){result[i]=getCount(i);}return result;}public int getCount(int num){int count=0;while(num!=0){if((num&1)==1){count++;}num/=2;}return count;}
}

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