求整数数组中和最大的子数组的3种方法

输入一个整型数组，数组里有正数也有负数。数组中连续的一个或多个整数组成一个子数组，每个子数组都有一个和。求所有子数组的和的最大值。要求时间复杂度为O(n)。

使用三种方式实现，分别是蛮力法：bf_MaxSubArray，分治法：dc_MaxArray，和O(n)实现的单循环法find_max。

#include"stdio.h"
#include"math.h"
#include"time.h"
#include"windows.h"
#define MIN_INT -10000
#define MAX_INT 10000
typedef struct{
	int max_left;
	int max_right;
	int max;
}maxinfo;


maxinfo bf_MaxSubArray(int *a,int n)
{
	maxinfo max;
	max.max_left= 0;
	max.max_right= 0;
	max.max= MIN_INT;
	int tmp=0;
	for(int i=0;i<n;i++)
	{
		for(int j=i;j<n;j++)
		{
			tmp += a[j];
			if(tmp > max.max)
			{
				max.max = tmp;
				max.max_left = i;
				max.max_right = j;
			}
		}
		tmp = 0;
	}

	return max;
}

maxinfo dc_FindMaxCrossingSubArray(int *a,int left,int mid,int right)
{
	maxinfo leftMax;
	leftMax.max_left= mid;
	leftMax.max_right = mid;
	leftMax.max= MIN_INT;
	int tmp = 0;
	for(int i=mid;i>=left;i--)
	{
		tmp += a[i];
		if(tmp>leftMax.max)
		{
			leftMax.max = tmp;
			leftMax.max_left = i;
		}
	}

	maxinfo rightMax;
	rightMax.max_left=mid+1;
	rightMax.max_right = mid+1;
	rightMax.max= MIN_INT;
	tmp = 0;
	for(int j = mid+1;j<right+1;j++)
	{
		tmp += a[j];
		if(tmp>rightMax.max)
		{
			rightMax.max = tmp;
			rightMax.max_right = j;
		}
	}

	leftMax.max += rightMax.max;
	leftMax.max_right = rightMax.max_right;
	return leftMax;

}
maxinfo dc_MaxArray(int *a,int left,int right)
{
	if(left==right)
	{
		maxinfo sgele;
		sgele.max_left=left;
		sgele.max_right = right;
		sgele.max= a[left];
		return sgele;
	}
	else
	{
		int mid = (int)floor((left+right)/2);
		maxinfo tmp;
		maxinfo maxleft = dc_MaxArray(a,left,mid);
		maxinfo maxright = dc_MaxArray(a,mid+1,right);
		maxinfo maxcross = dc_FindMaxCrossingSubArray(a,left,mid,right);
		tmp = maxleft.max>maxright.max?maxleft:maxright;
		tmp = tmp.max>maxcross.max?tmp:maxcross;
		return tmp;
	}
}
void show(maxinfo max)
{
	printf("收益最大的是：%d号前买入，%d号后卖出，收益$%d/股.\n",max.max_left+1,max.max_right+1,max.max);
}

void find_max(int *ary,int ARYSIZE)
{
	int max=0;//保存最大和
	int curSum=0;//保存当前和
	int curStart=0;//当前和的起始位置
	int start=0;//最大和的起始位置
	int end=0;//最大和的终止位置
	for(int i=0;i<ARYSIZE;i++)
	{
		if(i==0)
		{
			curSum=max=ary[i];
			continue;
		}
			
		if(curSum<0)
		{
			curSum=0;//与负数相加，和会减小，所以抛弃以前的和
			curStart=i;
		}
		//最大值已经被保存下来，所以请大胆的继续往前加
		curSum += ary[i];
		//当前和被保存为最大值，记录下它的起始位置和结束位置
		if(curSum>max)
		{
			max=curSum;
			start=curStart;
			end=i;
		}
	}

	printf("和最大的子数组为：\n");
	for(int j=start;j<=end;j++)
	{
		printf("%d ",ary[j]);
	}
	printf("= %d",max);
}
void main()
{
	int n = 16;
	int a[] = {13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7};
	//{-13,-3,-25,-20,-3,-16,-23,-18,-20,-7,-12,-5,-22,-15,-4,-7};//
	int i=clock();
	show(bf_MaxSubArray(a,n));
	printf("\n==============\n");

	find_max(a,n);
	printf("\n==============\n");
	printf("计算耗时:%dms\n",clock()-i);

	int mid = (int)floor((n-1)/2);
	show(dc_FindMaxCrossingSubArray(a,0,mid,n-1));


	maxinfo result;
	LARGE_INTEGER nFreq;

	LARGE_INTEGER nBeginTime;

	LARGE_INTEGER nEndTime;

	double time;

	QueryPerformanceFrequency(&nFreq);

	QueryPerformanceCounter(&nBeginTime); 
	//开始
	result = bf_MaxSubArray(a,n);//dc_MaxArray(a,0,n-1);

	QueryPerformanceCounter(&nEndTime);

	time=(double)(nEndTime.QuadPart-nBeginTime.QuadPart)/(double)nFreq.QuadPart;
	
	printf("%f微妙\n",time);
	show(result);
	printf("\n");  
	
}

其中最优方法的参考地址：求整数数组中和最大的子数组