LeetCode 338. Counting Bits

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.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

【题目分析】

给定一个数num，返回0 ≤ i ≤ num的所有数的二进制表示形式中包含的‘1’的个数。

【思路】

1. 时间复杂度为O(n*sizeof(integer))的方法

对于0到num中的每个数，如果这个数是奇数，则1的个数加1，然后该数除以2，直到该数为0. 代码如下：

 1 public class Solution {
 2     public int[] countBits(int num) {
 3         int[] result = new int[num+1];
 4         for(int i = 0; i <= num; i++) {
 5             int count = 0, x = i;
 6             while(x != 0) {
 7                 if(x % 2 == 1) count++;
 8                 x = x >> 1;
 9             }
10             result[i] = count;
11         }
12         return result;
13     }
14 }

2. 时间复杂度为O(N)，空间复杂度为O(N)

求解过程中求出的一些结果其实可以直接用在后面求解中，可以大大节省代码时间复杂度，代码如下：

 1 public class Solution {
 2     public int[] countBits(int num) {
 3         int[] result = new int[num+1];
 4         for(int i = 1; i <= num; i++) {
 5             if(i % 2 == 1) {
 6                 result[i] = result[i>>1] + 1;
 7             } else {
 8                 result[i] = result[i>>1];
 9             }
10         }
11         return result;
12     }
13 }

3.对上面的代码做进一步的优化

1 public int[] countBits(int num) {
2     int[] f = new int[num + 1];
3     for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
4     return f;
5 }

java的位运算符要比算术运算符快很多