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 ", result, utimecost); // adopt algorithm 2 utimecost = GetTickCount(); result = maxsubsum2(ptr, n); utimecost = GetTickCount() - utimecost; printf("max subsequence sum is %d, time cost %d ", result, utimecost); free(ptr); return 0; } |
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
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/*************************************************************************
> 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 ", result, utimecost); // adopt algorithm 3 utimecost = GetTickCount(); result = maxsubsum3(ptr, n); utimecost = GetTickCount() - utimecost; printf("max subsequence sum is %d, time cost %d ", result, utimecost); free(ptr); return 0; } |
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。
