Leetcode 解題報告

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338. Counting Bits --20160518

 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?

思路：這是一個找規律的題，前後數位1的個數是有增長的規律的。在稿紙上寫出來就可以清晰地看到，這裡就不贅述，直接上代碼。

public class S338 {    public int[] countBits(int num) {        int[] count = new int[num+1];        int k = 0;        for (int i = 1; i <= num;) {            int temp = (int)Math.pow(2, k);            for (int j = 0; j < temp; j++) {                count[i] = count[i - temp] + 1;                i++;                if(i > num) {                    break;                }            }            k++;        }        return count;    }}

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Leetcode 解題報告