8.旋转数组的最小数

#include <iostream>

using namespace std;


//int nums[]{1, 0, 1, 1, 1};    无法确定中间的数字1属于第一个递增的子序列还是第二个递增的子序列
int BinMin(int* numbers, int length)
{
    if(numbers == nullptr || length <= 0)
        return 0x80000000;  //最小的负数

    int index1 = 0;
    int index2 = length - 1;
    int indexMid = index1;

    while(numbers[index1] >= numbers[index2])
    {
        if(index2-index1 == 1)
        {
            indexMid = index2;
            break;
        }

        indexMid = (index1+index2)/2;
        if(numbers[indexMid] >= numbers[index1])
        {
            index1 = indexMid;
        }
        else if(numbers[indexMid] <= numbers[index2])
        {
            index2 = indexMid;
        }
    }

    return numbers[indexMid];
}



int Min(int* numbers, int length)
{
    if(numbers == nullptr || length <= 0)
        return 0x80000000;  //最小的负数

    int index1 = 0;
    int index2 = length - 1;
    int indexMid = index1;

    while(numbers[index1] >= numbers[index2])
    {
        if(index2-index1 == 1)
        {
            indexMid = index2;
            break;
        }

        indexMid = (index1+index2)/2;

        // 如果index1，index2，indexmid 指向的数相等，只能采用顺序查找
        if(numbers[index1] == numbers[index2] && numbers[indexMid] == numbers[index1])
        {
            int result = numbers[index1];
            for(unsigned i = index1+1; i <= index2; i++)
            {
                if(numbers[i] < result)
                {
                    result = numbers[i];
                }
            }

            return result;
        }

        if(numbers[indexMid] >= numbers[index1])
        {
            index1 = indexMid;
        }
        else if(numbers[indexMid] <= numbers[index2])
        {
            index2 = indexMid;
        }
    }

    return numbers[indexMid];
}

int main()
{

    //int nums[]{3, 4, 5, 1, 2};
    int nums[]{1, 0, 1, 1, 1};
    cout<<Min(nums, sizeof(nums)/sizeof(nums[0]))<<endl;
    return 0;
}