Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Example
Example 1:

Input: [4, 5, 1, 2, 3] and target=1, 
Output: 2.
Example 2:

Input: [4, 5, 1, 2, 3] and target=0, 
Output: -1.

Author: SarahChenJieyi

Search in Rotated Sorted Array

@@ -0,0 +1,31 @@
+class Solution:
+    """
+    @param A: an integer rotated sorted array
+    @param target: an integer to be searched
+    @return: an integer
+    """
+    def search(self, A, target):
+        # write your code here
+        if not A:
+            return -1
+        
+        start, end = 0, len(A) - 1
+        while start+1 < end:
+            mid = (start + end) // 2
+            if A[mid] > A[start]:
+                if A[start] <= target <= A[mid]:
+                    end = mid
+                else:
+                    start = mid
+            else:
+                if A[mid] <= target <= A[end]:
+                    start = mid
+                else:
+                    end = mid
+        
+        if A[start] == target:
+            return start
+        elif A[end] == target:
+            return end
+        else:
+            return -1