Java [Leetcode 338]Counting Bits

时间:2024-07-29 20:05:20

题目描述:

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.

Hint:

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

解题思路:

0     0000

1     0001

2     0010

3     0011

4     0100

5     0101

6     0110

7     0111

8     1000

9     1001

10   1010

11   1011

12   1100

13   1101

14   1110

15   1111

........

观察上面的情况,我们发现0,1,2-3,4-7,8-15为一组,且每组开头的1的位数都是1。每组其余的数都可以用本组开头的数加上另一个差值。且这两个数都已经在前面算过了。

代码如下:

public class Solution{
public int[] countBits(int num){
int[] res = new int[num + 1];
int pow = 1, k = 1;
res[0] = 0;
while(k <= num){
if(k == pow){
pow *= 2;
res[k++] = 1;
} else {
res[k] = res[pow / 2 ] + res[k - pow / 2];
k++;
}
}
return res;
}
}