最大子数组

该博客探讨了一个C++程序,用于计算一维数组中具有最大和的连续子数组。程序首先生成随机数并写入文件,然后读取这些数并应用分治策略的动态规划算法`maxSubArray`来找到最大子数组和。实验中还包含了性能测试,以展示算法在大量数据上的运行效率。

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#include <iostream>
#include <cmath>
#include <Windows.h>
#include <stdlib.h>
#include <fstream>
#include <time.h>
using namespace std;


int nums[12] = { 1,-2,4,5,-2,8,3,-2,6,3,7,-1 };
bool randNums(int n) {
	fstream ofs("data.txt", ios::out | ios::trunc);
	if (!ofs.is_open()) {
		cout << "文件打开失败!" << endl;
		return false;
	}
	ofs << n << ' ';
	int temp = 0;
	for (int i = 0; i < n; i++) {
		temp = rand() % 200 - 100;  //随机数在-100到100
		ofs << temp << ' ';
	}
	return true;
}

int crossingSubArray(int nums[], int low, int mid, int height)
{

	int sum = 0, sLeft = INT_MIN, sRight = INT_MIN;

	//左边
	for (int i = mid; i >= low; i--) 
	{
		sum += nums[i];
		if (sum > sLeft) 
		{ //
			sLeft = sum;
		}
	}
	//右边
	sum = 0;
	for (int i = mid+1; i <= height; i++) 
	{
		sum += nums[i];
		if (sum > sRight) 
		{ //
			sRight = sum;
		}
	}
	return (sLeft > sRight ? sLeft : sRight) > (sRight + sLeft) ? 
		(sLeft > sRight ? sLeft : sRight) : (sRight + sLeft);
}

int maxSubArray(int nums[],int low,int height)
{
	if (low == height)  //分成最后一个元素
	{
		return nums[low];
	}

	int mid = (int)floor ((height + low) / 2);
	int S1 = maxSubArray(nums, low, mid);
	int S2 = maxSubArray(nums, mid + 1, height);

	int S3 = crossingSubArray(nums, low, mid, height);

	return ((S1 > S2) ? S1 : S2) > S3 ? ((S1 > S2) ? S1 : S2) : S3;

}	
int* readNums() {
	fstream ifs("data.txt", ios::in);
	int n;
	ifs >> n;
	int i = 0;
	int* nums = new int[n];
	while (ifs >> n) {
		//cout << n << ' ';
		nums[i++] = n;
	}
	cout << endl;
	cout << endl;
	return nums;
}
int main() 
{
	int n = 1000000;  //次数修改  建议次数大于1000000+才有效果
	srand((unsigned int)time(NULL));
	if (!randNums(n)) {
		return 0;
	}
	int* nums = readNums();

	clock_t sTime, eTime;
	sTime = clock();
	ULONGLONG startTime= GetTickCount64();
	cout << maxSubArray(nums, 0, n-1) << endl;
	ULONGLONG endTime = GetTickCount64();
	eTime = clock();
	cout << "THe run time is:" << endTime - startTime << "ms" << endl;
	cout << "THe run time is:" << (double)(eTime - sTime) / CLOCKS_PER_SEC << "s" << endl;

	return 0;
}

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