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.

Analysis:

Suppose we have three pointers:start, mid and end. There are three case for the array between start and end:

1. A[mid] < A[start]: Rotated, the maximum value is on the left of mid. If target < A[mid], then we search the left part; if target > A[mid], we then have two situations: (a) target <= A[end], we search the right part; (b) target > A[end], search the left part.

2. A[mid] > A[end]: Rotated, the maximum value is on the right of mid. target > A[mid]: search the left part; target < A[mid] && < A[start]: search the right part; target < A[mid] && >= A[start]: search the left part.

3. Otherwise: Not rotated. Just perform normal binary search.

Solution:

 1 public class Solution { 2     public int search(int[] A, int target) { 3         if (A.length==0) return -1; 4  5         int start = 0, end = A.length-1; 6         int mid = -1; 7  8         while (start<=end){ 9             mid = (start+end)/2;10             if (A[mid]==target) return mid;11 12             if (A[mid]<A[start]){13                 if (target<A[mid])14                     end = mid-1;15                 else if (target<=A[end])16                     start = mid+1;17                 else end = mid-1;18                 continue;19             }20 21             if (A[mid]>A[end]){22                 if (target>A[mid]) start = mid+1;23                 else if (target>=A[start]) end = mid-1;24                 else start = mid+1;25                 continue;26             }27 28             if (target<A[mid]) end = mid-1;29             else start = mid+1;30         }31 32         return -1;33     }34 }

Leetcode-Search in Rotated Sorted Array