求旋转数组的最小数字

题目：把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如数组{3, 4, 5, 1, 2}为{1, 2, 3, 4, 5}的一个旋转，该数组的最小值为1。

#include<iostream>
using namespace std;
//顺序查找
int MinInOrder(int*numbers, int index1, int index2)
{
	int result = numbers[index1];
	for (int i = index1 + 1;i <= index2;++i)
	{
		if (result > numbers[i])
			result = numbers[i];
	}
	return result;
}

int MIn(int* numbers, int length)
{
	if (numbers == nullptr&&length <= 0)
		throw new exception("Invalid Parater");
	int index1 = 0;
	int index2 = length - 1;
	int indexMid = index1;
	while (numbers[index1] >= numbers[index2])
	{
//如果index1和index2指向相邻的两个数，则index1指向第一个递增子数组的最后一个数字，
//index2指向第二个子数组的第一个数字，也就是数组中的最小数字
		if (index2 - index1 == 1)
		{
			indexMid = index2;
			break;
		}
		int indexMid = (index1 + index2) / 2;
//如果下标为index1、index2和indexMid指向的三个数字相等，则只能顺序查找
		if (numbers[index1] == numbers[index2] && numbers[index1] == numbers[indexMid])
			return MinInOrder(numbers, index1, index2);
		if (numbers[index1] <= numbers[indexMid])
			index1 = indexMid;
		if (numbers[index2] >= numbers[indexMid])
			index2 = indexMid;
	}
	return numbers[indexMid];
}
//测试输入
void Test1()
{
	int numbers[] = {3,4,5,1,2};
	printf("旋转数组的最小数是：%d\n",MIn(numbers,sizeof(numbers)/sizeof(numbers[0])));
}

void Test2()
{
	int numbers[] = { 3, 4, 5, 1, 1, 2 };
	printf("旋转数组的最小数是：%d\n", MIn(numbers, sizeof(numbers) / sizeof(numbers[0])));
}

void Test3()
{
	int numbers[] = { 1, 2, 3, 4, 5 };
	printf("旋转数组的最小数是：%d\n", MIn(numbers, sizeof(numbers) / sizeof(numbers[0])));
}

void Test4()
{
	printf("旋转数组的最小数是：%d\n", MIn(nullptr, 0));
}

int main()
{
	Test1();
	Test2();
	Test3();
	Test4();
	return 0;
}