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LeetCode_338_CountingBits

LeetCode_338_CountingBits

作者: 水月心刀 | 来源:发表于2016-11-03 20:13 被阅读11次

    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.

    题目分析

    对于一个给定的数字num, 给出[0,num]区间内的每个数的二进制表示所含的1的个数,用一个数组输出结果。
    尝试给出符合以下条件的solution

    1. T(n) = O(n)
    2. S(n) = O(n)
    3. 不使用C++或其他语言的内置函数

    解:
    假定rst[num] 即为num的二进制表示法中包含的1的个数,存在以下公式
    rst[num] = rst[num^(num-1)] + 1
    且rst[0] = 0;

    Solution:

    vector<int> countBits(int num)
    {
       vector<int> ret(num + 1, 0);
       for (int i = 1; i <= num; ++i)
           ret[i] = ret[i&(i - 1)] + 1;
       return ret;
    }
    

    T(n) = O(n)
    S(n) = O(n)

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