算法：Solutions for the Maximum Subsequence Sum Problem

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最新推荐文章于 2024-03-30 20:15:57 发布

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分类专栏：数据结构与算法文章标签： maximum sub array

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本文对比了四种不同算法在解决最大子数组求和问题上的性能表现，包括两种基于遍历的方法、一种分治法及一种线性时间复杂度算法。通过对不同规模数据集的实验，展示了各种算法的实际运行效率。

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The maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array of numbers (containing at least one positive number) which has the largest sum. For example, for the sequence of values −2, 1, −3, 4, −1, 2, 1, −5, 4; the contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6. --from wiki

下面我们分析四种算法的时间性能,由于运行时间相差较大，我们分成两组进行对比：

环境：ubuntu 12.04

时间单位：ms

时间性能：presume that the input is preread

第一组：输入数据元素个数2000

    C++ Code 
 

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/*************************************************************************
    > File Name: algorithm1.c
    > Author: Simba
    > Mail: dameng34@163.com
    > Created Time: 2012年12月24日星期一 22时41分56秒
************************************************************************/

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<sys/time.h>

int maxsubsum1( const int a[], int n)
{
     int thissum, maxsum, i, j, k;

    maxsum = 0;
     for (i = 0; i < n; i++)
    {
         for (j = i; j < n; j++)
        {
            thissum = 0;
             for (k = i; k <= j; k++)
                thissum += a[k];

             if (thissum > maxsum)
                maxsum = thissum;
        }
    }
     return maxsum;
}

int maxsubsum2( const int a[], int n)
{
     int thissum, maxsum, i, j;

    maxsum = 0;
     for (i = 0; i < n; i++)
    {
        thissum = 0;
         for (j = i; j < n; j++)
        {
            thissum += a[j];

             if (thissum > maxsum)
                maxsum = thissum;
        }
    }
     return maxsum;
}

long GetTickCount( void)
{
     struct timeval tv;

    gettimeofday(&tv, NULL);

     return (tv.tv_sec * 1000 + tv.tv_usec / 1000);
}

int main( void)
{
     int i, n = 2000;
     int *ptr = malloc( sizeof( int) * n);
    srand(time( NULL));
     for (i = 0; i < n; i++)
        ptr[i] = rand() % 50 - 25;
     // adopt algorithm  1
     unsigned int utimecost = GetTickCount();
     int result = maxsubsum1(ptr, n);
    utimecost = GetTickCount() - utimecost;
    printf( "max subsequence sum is %d, time cost %d\n", result, utimecost);

     // adopt algorithm  2
    utimecost = GetTickCount();
    result = maxsubsum2(ptr, n);
    utimecost = GetTickCount() - utimecost;
    printf( "max subsequence sum is %d, time cost %d\n", result, utimecost);

    free(ptr);

     return 0;
}

输出为：

max subsequence sum is 275, time cost 4423
max subsequence sum is 275, time cost 6

第二组：输入数据元素个数 1000000

    C++ Code 
 

/*************************************************************************
    > File Name: divide_conquer.c
    > Author: Simba
    > Mail: dameng34@163.com
    > Created Time: 2012年12月24日星期一 23时24分41秒
************************************************************************/

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include <sys/time.h> /* struct timeval, gettimeofday(), struct itimerval, setitimer(), ITIMER_REAL */

int divide_conquer( int arr[], int start, int end)
{
     if(start == end)
         return (arr[start] > 0 ? arr[start] : 0);

     int mid = (start + end) / 2;
     int max_left = divide_conquer(arr, start, mid);
     int max_right = divide_conquer(arr, mid + 1, end);
     // mid subsequence

     int max_left_border = 0;
     int tmp_sum = 0;
     int i;

     for(i = mid; i >= start; i--)
    {
        tmp_sum += arr[i];
         if(tmp_sum > max_left_border)
            max_left_border = tmp_sum;
    }

     int max_right_border = 0;
    tmp_sum = 0;
     for(i = mid + 1; i <= end; i++)
    {
        tmp_sum += arr[i];
         if(tmp_sum > max_right_border)
            max_right_border = tmp_sum;
    }

     int max_mid = max_left_border + max_right_border;
     // max subsequence
     int iresult = max_left;
     if(max_right > iresult)
        iresult = max_right;
     if(max_mid > iresult)
        iresult = max_mid;
     return iresult;
}

int maxsubsum3( const int a[], int n)
{
     int j, thissum, maxsum;
    thissum = maxsum = 0;
     for (j = 0; j < n; j++)
    {
        thissum += a[j];

         if (thissum > maxsum)
            maxsum = thissum;
         else if (thissum < 0)
            thissum = 0;
    }

     return maxsum;
}

long GetTickCount( void)
{
     struct timeval tv;

    gettimeofday(&tv, NULL);

     return (tv.tv_sec * 1000 + tv.tv_usec / 1000);
}

int main( void)
{
     int i, n = 1000000;
     int *ptr = malloc( sizeof( int) * n);
    srand(time( NULL));
     for (i = 0; i < n; i++)
        ptr[i] = rand() % 50 - 25;
     // adopt divide_conquer algorithm
     unsigned int utimecost = GetTickCount();
     int result = divide_conquer(ptr, 0, n - 1);
    utimecost = GetTickCount() - utimecost;
    printf( "max subsequence sum is %d, time cost %d\n", result, utimecost);
     // adopt algorithm 3
    utimecost = GetTickCount();
    result = maxsubsum3(ptr, n);
    utimecost = GetTickCount() - utimecost;
    printf( "max subsequence sum is %d, time cost %d\n", result, utimecost);

    free(ptr);

     return 0;
}

输出为：

max subsequence sum is 2410, time cost 217
max subsequence sum is 2410, time cost 4

分析：

在《data structure and algorithm analysis in c》中有对这四种算法时间性能的分析，依次下来分别是O(n^3)，O(n^2)，O(nlogn)，O(n)，即使我们在第二组输入的元素个数是第一组的500倍，第二组的运行时间都要比第一组的小。下图2-2是作者写书时测试的时间列表，显然现在的机器运行得更快。