求最大子序列和的两种算法

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/*

*author: Jack.xu/nanac.xu

*date: 08/26/2009

*function: max subsequence sum

*/

#include

#include

namespace dskit{

/*with O(N) time complexity*/

int MSS(const ELEMENTSTYPE* elements, size_t size)

{

assert(NULL != elements);

    int the_sum = 0;

    int max_sum = 0;

    for(int i = 0; i != size; ++i)

    {

        the_sum += elements[i];

        if(the_sum > max_sum)

            max_sum = the_sum;

        else if(the_sum < 0)

            the_sum = 0;

    }

    return max_sum;

}

/*divided algorithm with O(NlogN) time complexity*/

int MSS_divide(const ELEMENTSTYPE* elements, size_t begin, size_t end)

{

    assert(NULL != elements && begin  >= 0 && end >= 0);

    if(begin == end)

    {

        if(elements[begin] < 0)

            return 0;

        else

            return elements[begin];

    }

    size_t center = ((begin + end) >> 1);

    int left_max = MSS_divide(elements, begin, center);

    int right_max = MSS_divide(elements, center + 1, end);

    int temp_sum = 0;

    int center_left_max = 0;

    for(int i = center; i >= begin; i--)

    {

        if(i < 0)

            break;

        temp_sum += elements[i];

        if(temp_sum > center_left_max)

            center_left_max = temp_sum;

    }

    temp_sum = 0;

    int center_right_max = 0;

    for(int i = center + 1; i <= end; i++)

    {

        temp_sum += elements[i];

        if(temp_sum > center_right_max)

            center_right_max = temp_sum;

    }

    temp_sum = (left_max > right_max) ? left_max : right_max;

    return (temp_sum > (center_right_max + center_left_max)) ? temp_sum : (center_right_max + center_left_max);

}

}

#ifndef NDEBUG

int main(int argc, char* argv[])

{

    ELEMENTSTYPE array[] = {-1, 1, 32, 3, 4, -23, 24};

    std::cout << dskit::MSS(array, sizeof(array) / sizeof(ELEMENTSTYPE)) << std::endl;

    std::cout << dskit::MSS_divide(array, 0, sizeof(array) / sizeof(ELEMENTSTYPE) - 1) << std::endl;

    return 0;

}

#endif

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