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  • LeetCode-Count 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.

    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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/5832068.html
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