Returns the smallest element after rotating an ordered array.

Sequential search can be found, but the time complexity is not the best.

Therefore, binary search can be considered.

In general, if the intermediate data is greater than the first data, it indicates that the intermediate data is a large part, and the smallest number is on the right. Modify the pointer to reloop. If the intermediate data is smaller than the last one, it indicates that the intermediate data is a small part, and the smallest number is on its left. Modify the pointer to reloop.

Some Boundary situations include:

1. No rotation at all

2. if 1 0 1 1 1 or 1 1 1 0 1, the size of the first data, intermediate data, and last data is the same, the smallest data may be on the left or right of the intermediate data, so it cannot be determined and can only be searched in sequence.

# Include <stdio. h> # include <stdlib. h> int search_min (int * data, int length) {int head = 0, result, I; int tail = length-1; int middle = head; // when the initialization is index1, if (data = NULL | length <= 0) return-1; while (data [head]> = data [tail]) {if (tail-head = 1) {middle = tail; break;} middle = (head + tail)/2; if (data [head] = data [tail] & data [head] = data [middle]) {// you can only search in sequence, in this case, it is easy to omit result = data [head + 1]; for (I = head + 2; I <= tail-1; ++ I) {if (result> data [I]) result = data [I];} return result;} if (data [middle]> = data [head]) head = middle; else if (data [middle] <= data [tail]) tail = middle;} return data [middle];} void main () {int d [5] = {1, 1, 3, 4, 5}; printf ("% d \ n", search_min (NULL, 5); getch ();}