旋转数组的最小数字——剑指offer

 1 #include<stdio.h>
 2 int MinInOrder(int r[], int p1, int p2)
 3 {
 4     int result = r[p1];
 5     for (int i = p1 + 1; i <= p2; i++)
 6     {
 7         if (r[i] < result)
 8             result = r[i];
 9     }
10     return result;
11 }
12 int Min(int r[], int length)
13 {
14     if (r == NULL || length <= 0)
15         return 0;
16     int p1 = 0;
17     int p2 = length - 1;
18     int mid = p1;
19     while (r[p1] >= r[p2])
20     {
21         if (p2 - p1 == 1)
22         {
23             mid = p2;
24             break;
25         }
26         mid = (p1 + p2) / 2;
27         if (r[p1] == r[p2] && r[mid] == r[p2])
28             return MinInOrder(r, p1, p2);
29         if (r[mid] >= r[p1])
30             p1 = mid;
31         else if (r[mid] <= r[p2])
32             p2 = mid;
33     }
34     return r[mid];
35 }
36 
37 int main()
38 {
39     int a[8] = { 1, 1, 1, 0, 1 };
40     printf("%d\n", Min(a, 5));
41     return 0;
42 }