首先利用二分查找的方法找到最小值，然后在以最小值为分界的两块分别进行二分搜索。

1 int findPos(int A[], int left, int right){ 2     if(left > right) 3         return -1; 4     int mid = (left+right)/2; 5     int result; 6     if(A[left] > A[mid]){ 7         result = findPos(A, left, mid-1); 8         if(result == -1) 9             return mid;10         else11             return A[result]
     
       right)23         return -1;24     int mid = (left+right)/2;25     if(A[mid] == target)26         return mid;27     else if(A[mid] < target)28         return bsearch(A, mid+1, right, target);29     else30         return bsearch(A, left, mid-1, target);31 }32     int search(int A[], int n, int target) {33         int pos = findPos(A, 0, n-1);34         int result = bsearch(A, 0, pos-1, target);35         if(result != -1)36             return result;37         return bsearch(A, pos, n-1, target);38     }
     

 另外一种方法是直接二分，不过增加了一些判断，比上述代码要简练。

1     int search(int A[], int n, int target) { 2         if(n == 0) 3             return -1; 4         int l = 0, r = n-1, m; 5         while(l <= r){ 6             m = (l+r)/2; 7             if(A[m] == target) 8                 return m; 9             else if(A[m] > A[l]){10                 if(target < A[m] && target >= A[l])11                     r = m-1;12                 else13                     l = m+1;14             }15             else if(A[m] < A[l]){16                 if(target > A[m] && target <= A[r])17                     l = m+1;18                 else19                     r = m-1;20             }21             else22                 l++;23         }24         return -1;25     }