LeetCode-Count Bits

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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.

Analysis:

0 | 0+1| 0+1 1+1| 0+1 1+1 1+1 2+1 |..........

0 | 1    | 1       2  | 1      2    2     3     |................

Solution:

public class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num + 1];

        res[0] = 0;
        int nextNum = 1;
        int nextCount = 1;
        while (nextNum <= num) {
            for (int i = 0; i < nextCount && nextNum <= num; i++) {
                res[nextNum++] = res[i] + 1;
            }
            nextCount *= 2;
        }
        return res;
    }
}

LeetCode-Count Bits

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原文地址：http://www.cnblogs.com/lishiblog/p/5832068.html