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# 62.Search in Rotated Sorted Array

## 1.Description(Medium)

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**

For`[4, 5, 1, 2, 3]`and`target=1`, return`2`.

For`[4, 5, 1, 2, 3]`and`target=0`, return`-1`.

[**Challenge**](https://www.lintcode.com/en/problem/search-in-rotated-sorted-array/#challenge)

O(logN) time

[**Tags**](https://www.lintcode.com/en/problem/search-in-rotated-sorted-array/#tags)

[Binary Search](https://www.lintcode.com/tag/binary-search/) [LinkedIn](https://www.lintcode.com/tag/linkedin/) [Array](https://www.lintcode.com/tag/array/) [Facebook](https://www.lintcode.com/tag/facebook/) [Sorted Array](https://www.lintcode.com/tag/sorted-array/) [Uber](https://www.lintcode.com/tag/uber/)

## 2.Code

注意分两种情况，看mid在pivot的哪边。

```
public int search(int[] A, int target) {
        if(A==null || A.length==0){
            return -1;
        }
        int left=0;
        int right=A.length-1;
        while(left+1<right){
            int mid=left+(right-left)/2;
            if(A[mid]==target){
                return mid;
            }
            //situation 1:
            if(A[left]>A[mid]){
                if(A[mid]<target && target<=A[right]){
                    left=mid;
                }
                else{
                    right=mid;
                }
            }
            //situation 2:
            if(A[left]<=A[mid]){
                if(A[left]<=target && target<A[mid]){
                    right=mid;
                }
                else{
                    left=mid;
                }
            }
        }
        if(A[left]==target){
            return left;
        }
        if(A[right]==target){
            return right;
        }
        return -1;
    }
```