求最大子序列及其效率解析

#include <iostream>
#include <vector>
using namespace std;

//s(tart)表示最大子序列的开始位置，e(nd)表示结束位置
//这里如果有多于一个的最大子序列的时候，只记录开始位置最低的那个
int s=0;
int e=0;

//穷举法，复杂度O(n^3)
long maxSubSum1(const vector<int> &a){
	long maxSum=0;
	for (int i=0; i<a.size();i++)
	{
		for (int j=i;j<a.size();j++)
		{
			long thisSum=0;
			for (int k=i; k<=j; k++)
			{
				thisSum+=a[k];
			}
			if (thisSum>maxSum){
				maxSum=thisSum;
				s=i;
				e=j;
			}
		}
	}
	return maxSum;
}

//也是穷举法，不过减去了上面的一些不必要操作O(n^2)  
long maxSubSum2(const vector<int> &a){
	long maxSum=0;
	for (int i=0; i<a.size(); i++)
	{
		long thisSum=0;
		for (int j=i; j<a.size(); j++)
		{
			thisSum+=a[j];
			if (thisSum>maxSum){
				maxSum=thisSum;
				s=i;
				e=j;
			}
		}
	}
	return maxSum;
}

long max3(long a, long b, long c){
	if(a<b)
		a=b;
	if(a>c)
		return a;
	else
		return c;
}

long maxSumRec(const vector<int> a, int left, int right){
	if(left == right)
	{
		//其实这个基准值在后面计算的时候可以保证
		//在这里不必多此一举
		if(a[left]>0)
			return a[left];
		else
			return 0;
	}

	int center=(left+right)/2;
	long maxLeftSum=maxSumRec(a,left,center);
	long maxRightSum=maxSumRec(a,center+1,right);

	//某段序列中，求含最右侧元素序列和的最大值
	long maxLeftBorderSum=0,leftBorderSum=0;
	for (int i=center; i>=left; i--)
	{
		leftBorderSum+=a[i];
		if(leftBorderSum>maxLeftBorderSum)
		{
			maxLeftBorderSum=leftBorderSum;
			s=i;
		}
	}
	//某段序列中，求含最左侧元素序列和的最大值
	long maxRightBorderSum=0,rightBorderSum=0;
	for (int j=center+1; j<=right; j++)
	{
		rightBorderSum+=a[j];
		if(rightBorderSum>maxRightBorderSum)
		{
			maxRightBorderSum=rightBorderSum;
			e=j;
		}
	}

	return max3(maxLeftSum,maxRightSum,
		maxLeftBorderSum+maxRightBorderSum);
}

//该方法我们采用“分治策略”（divide-and-conquer）,相对复杂的O(NlogN)的解法
//最大子序列可能在三个地方出现，或者在左半部，或者在右半部，
//或者跨越输入数据的中部而占据左右两部分。前两种情况递归求解，
//第三种情况的最大和可以通过求出前半部分最大和（包含前半部分最后一个元素）
//以及后半部分最大和（包含后半部分的第一个元素）相加而得到。
long maxSubSum3(const vector<int> &a){
	return maxSumRec(a,0,a.size()-1);
}

//如果a[i]是负数那么它不可能代表最有序列的起点，因为任何包含a[i]的作为起点的子
//序列都可以通过用a[i+1]作为起点来改进。类似的有，任何的负的子序列不可能是最优
//子序列的前缀。例如说，循环中我们检测到从a[i]到a[j]的子序列是负数，那么我们就可以推进i。
//关键的结论是我们不仅可以把i推进到i+1，而且我们实际可以把它一直推进到j+1。
long maxSubSum4(const vector<int> &a){
	long maxSum=0;
	long thisSum=0;
	int t=0;
	for (int j=0; j<a.size(); j++)
	{
		thisSum+=a[j];
		if(thisSum>maxSum){
			maxSum=thisSum;
			s=t;
			e=j;
		}
		else if(thisSum<0){
			thisSum=0;
			t=j+1;
		}
	}
	return maxSum;
}

测试程序：

#include <iostream>
#include <vector>
#include <string>
#include <fstream>
#include <ctime>
#include <cstdlib>
using namespace std;

const long COUNT = 1000;
const int MAX_NUM = 200;
int Start = 0;
int End = 0;

bool readFile(vector<int>&input, string fileName)
{
	ifstream infile(fileName.c_str());
	if(!infile)
	{
		return false;
	}
	int s;
	while(infile >> s)
	{
		input.push_back(s);
	}
	return true;
}

bool writeTestData(string fileName)
{
	ofstream outfile(fileName.c_str());
	if(!outfile)
	{
		return false;
	}
	srand((unsigned)time(NULL));
	for(int i = 0; i < COUNT; i++)
	{
		if(rand() % 2 == 0)
		{
			outfile << rand() % MAX_NUM << endl;
		}
		else 
		{
			outfile << ~(rand() % MAX_NUM) <<endl;
		}
	}
	return true;
}
long maxSubSum1(const vector<int>& a)
{
	long maxSum = 0;
	for(int i = 0; i < a.size(); i++)
	{
		for(int j = i; j < a.size();j++)
		{
			long thisSum = 0;
			for(int k = i; k <= j; k++)
			{
				thisSum += a[k];
			}
			if(thisSum > maxSum)
			{
				maxSum = thisSum;
				Start = i + 1;
				End = j + 1;
			}
		}
	}
	return maxSum;
}

long maxSubSum2(const vector<int>& a)
{
	long maxSum = 0;
	for(int i = 0; i < a.size(); i++)
	{
		long thisSum = 0;
		for(int j = i; j <a.size(); j++)
		{
			thisSum += a[j];
			if(thisSum > maxSum)
			{
				maxSum = thisSum;
				Start = i + 1;
				End = j + 1;
			}
		}
	}
	return maxSum;
}

long max3(long a, long b, long c)
{
	if(a < b)
	{
		a = b;
	}
	if(a > c)
	{
		return a;
	}
	else
	{
		return c;
	}
}

long maxSumRec(const vector<int>&a, int left, int right)
{
	if(left == right)
	{
		if(a[left] > 0)
		{
			return a[left];
		}
		else return 0;
	}
	int center = (left + right) / 2;
	long maxLeftSum = maxSumRec(a, left, center);
	long maxRightSum = maxSumRec(a, center + 1, right);

	long maxLeftBorderSum = 0, leftBorderSum = 0;
	for (int i = center; i >= left; i--)
	{
		leftBorderSum += a[i];
		if (leftBorderSum > maxLeftBorderSum)
		{
			maxLeftBorderSum = leftBorderSum;
			Start = i + 1;
		}
	}

	long maxRightBorderSum = 0, rightBorderSum = 0;
	for(int j = center + 1; j <= right; j++)
	{
		rightBorderSum += a[j];
		if(rightBorderSum > maxRightBorderSum)
		{
			maxRightBorderSum = rightBorderSum;
			End = j + 1;
		}
	}
	return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}

long maxSubSum3(const vector<int>&a)
{
	return maxSumRec(a, 0, a.size() - 1);
}

long maxSubSum4(const vector<int>& a)
{
	long maxSum = 0, thisSum = 0;
	for (int j = 0; j < a.size(); j++)
	{
		thisSum += a[j];
		if (thisSum > maxSum)
			maxSum = thisSum;
		else if (thisSum < 0)
			thisSum = 0;
	}
	return maxSum;
}

int main()
{
	clock_t start,finish;
	vector<int>num;
	if(!writeTestData("in.txt"))
	{
		cout << "写入文件错误" << endl;
	}
	if(readFile(num, "in.txt"))
	{
		start = clock();
		cout << maxSubSum1(num) << endl;
		finish = clock();
		//cout << "Start position = " << Start << "\t" << "Start position = " << End <<endl;
		printf("Time used = %.2lf\n",(double)(finish - start)/CLOCKS_PER_SEC); 

		start = clock();
		cout << maxSubSum2(num) << endl;
		finish = clock();
		//cout << "Start position = " << Start << "\t" << "Start position = " << End <<endl;
		printf("Time used = %.2lf\n",(double)(finish - start)/CLOCKS_PER_SEC); 

		start = clock();
		cout << maxSubSum3(num) << endl;
		finish = clock();
		printf("Time used = %.2lf\n",(double)(finish - start)/CLOCKS_PER_SEC); 

		start = clock();
		cout << maxSubSum4(num) << endl;
		finish = clock();
		printf("Time used = %.2lf\n",(double)(finish - start)/CLOCKS_PER_SEC); 
	}
	return 0;
}

count = 1000

4135
Time used = 12.87
4135
Time used = 0.06
4135
Time used = 0.00
4135
Time used = 0.00
请按任意键继续. . .

count = 2000

8968
Time used = 104.37
8968
Time used = 0.23
8968
Time used = 0.00
8968
Time used = 0.00
请按任意键继续. . .

count = 10000 //O(N^3)略去

15302
Time used = 5.88
15302
Time used = 0.02
15302
Time used = 0.00
请按任意键继续. . .

count = 100000

33853
Time used = 570.03
33853
Time used = 0.16
33853
Time used = 0.01
请按任意键继续. . .

!!!