Rotate the smallest number in an array

Title Description: moves a number of elements at the beginning of an array to the end of an array, called a rotation of an array. Enter the rotation of an incrementally sorted array, outputting the smallest element of the rotated array.

For example: The array {3,4,5,1,2} is a rotation of {1,2,3,4,5}, and the smallest element of the array is 1.

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Int min (int* numbers, int length) {    if (Numbers == NULL  | |  length <= 0) &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;THROW&NBSP;NEW&NBSP;STD:: Exception ("Invalid parameters");     int index1 = 0;     int index2 = length - 1;    int indexmid =  index1;    while (Numbers[index1] >= numbers[index2])      {        //  if Index1 and Index2 point to adjacent two numbers,         //  index1 points to the last digit of the first incrementing Subarray,         // index2 points to the first number of the second Subarray, which is the smallest number         if in the array (index2 - &NBSP;INDEX1&NBSP;==&NBSP;1)         {             indexmid = index2;             break;        }          //  if the subscript is index1, Index2, and Indexmid point to a three number equal,        //  You can only search in sequence                  indexmid =  (INDEX1&NBSP;+&NBSP;INDEX2)  / 2;         //  narrowing the look-up range         if (numbers[indexmid] >=  NUMBERS[INDEX1])             index1 =  Indexmid;        else if (numbers[indexmid] <=  NUMBERS[INDEX2])             index2 =  indexmid;    }      return numbers[indexmid];} 

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Int min (int* numbers, int length) {    if (Numbers == NULL  | |  length <= 0) &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;THROW&NBSP;NEW&NBSP;STD:: Exception ("Invalid parameters");     int index1 = 0;     int index2 = length - 1;    int indexmid =  index1;    while (Numbers[index1] >= numbers[index2])      {        //  if Index1 and Index2 point to adjacent two numbers,         //  index1 points to the last digit of the first incrementing Subarray,         // index2 points to the first number of the second Subarray, which is the smallest number         if in the array (index2 - &NBSP;INDEX1&NBSP;==&NBSP;1)         {             indexmid = index2;             break;        }          //  if the subscript is index1, Index2, and Indexmid point to a three number equal,        //  Only sequential lookup         indexMid =  (INDEX1&NBSP;+&NBSP;INDEX2)  /  2;        if (numbers[index1] == numbers[index2]  &AMP;&AMP;&NBSP;NUMBERS[INDEXMID]&NBSP;==&NBSP;NUMBERS[INDEX1])              return mininorder (NUMBERS,&NBSP;INDEX1,&NBSP;INDEX2);         //  narrowing the look-up range         if (numbers[ INDEXMID]&NBSP;&GT;=&NBSP;NUMBERS[INDEX1])              index1 = indexmid;&Nbsp;       else if (Numbers[indexmid] <= numbers[index2])             index2 = indexMid;     }     return numbers[indexmid];} 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;}

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Rotate the smallest number in an array