算法：Solutions for the Maximum Subsequence Sum Problem
算法：Solutions for the Maximum Subsequence Sum Problem
s1mba
发布于 2018-01-03 17:48:42
发布于 2018-01-03 17:48:42
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
/*************************************************************************
> 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
/*************************************************************************
> 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。
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原始发表：2013-04-27 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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