分治法—最大子段问题

#include<stdio.h>
#include<stdlib.h>
//分治法
//最大子段问题
int Maxsubsum(int* Array, int left, int right)
{
	int sum = 0,i;
	//当array只有一个元素的时候
	if (left == right)
	{
		if (Array[left] > 0) {
			sum = Array[left];
		}
		else
		{
			sum = 0;
		}
	}
	else
	{
		int center = (left + right) / 2;
		int leftsum = Maxsubsum(Array, left, center);
		int rightsum = Maxsubsum(Array, center + 1, right);
		int s1 = 0;
		int lefts = 0;
		for (i = center;i >= left; i--)
		{
			lefts = lefts + Array[i];
			if (lefts>s1)
			{
				s1 = lefts;
			}
		}
		int s2 = 0;
		int rights = 0;
		for (i = center + 1; i <= right; i++)
		{
			rights = rights + Array[i];
			if (rights>s2)
			{
				s2 = rights;
			}
		}
		sum = s1 + s2;//最大子段问题的第三种情况
		
		//第一种情况
		if (sum < leftsum)
		{
			sum = leftsum;
		}
		//第二种情况
		if (sum < rightsum)
		{
			sum = rightsum;
		}
	}
	return sum;
}
int main()
{
	int* Array = (int*)malloc(6 * sizeof(int));
	Array[0] = -2;
	Array[1] = 11;
	Array[2] = -4;
	Array[3] = 13;
	Array[0] = -5;
	Array[0] = -2;
  	int result = Maxsubsum(Array, 0, 5);
	printf("%d", result);
	return 0;
}

时间复杂度为：O（nlogn）