旋转数组最小数字

[code]int Min(int* numbers, int length)
{
    if (numbers == NULL || length < 0)
    {
        throw invalid_argument("Invalid parameters");
    }
    int low = 0;
    int high = length - 1;
    int mid = 0;  //如果是原序列

    while (numbers[low] >= numbers[high])
    {
        // 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 (high - low == 1)
        {
            mid = high;
            break;
        }

        mid = (low + high) / 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[mid] == numbers[low] && numbers[mid] == numbers[high])
        {
            numbers[mid] = MinInOrder(numbers, low, high);
            break;
        }
        if (numbers[mid] >=  numbers[low])
            low = mid;
        else if(numbers[mid] <= numbers[high])
        {
            high = mid;
        }
    }

    return numbers[mid];
}

int MinInOrder(int* numbers, int low, int high)
{
    int result = numbers[0];
    for (int i = low+1; i <= high; i++)
    {
        if (numbers[i] < result)
        {
            result = numbers[i];
        }
    }

    return result;
}