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  • 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 1:

    Input: 2
    Output: [0,1,1]

    Example 2:

    Input: 5
    Output: [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.

    这题妙在

    十进制:             0 1 2 3      4 5 6 7    8 9 10 11 12 13 14 15    16... 

    二进制中1的个数:0 1 1 2       1 2 2 3    1  2 2  3  2  3  3   4     1...

    这题最妙在于当遇到2的pow时,把t置零,然后在前面数字二进制中1的个数的基础上加1

    public class Solution {
        public int[] countBits(int num) {
            int[] res = new int[num+1];
            int pow = 1;
            res[0] = 0;
            for(int i=1,t=0;i<=num;i++,t++)
            {
                if(i==pow)
                {
                    pow = pow * 2;
                    t = 0;
                }
                res[i] = res[t] + 1;
            }
            return res;
            
        }
    }
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  • 原文地址:https://www.cnblogs.com/hygeia/p/9758094.html
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