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# Search in Rotated Sorted Array

题目：

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.

分析：

判断Mid处于上半段还是下半段后，在单调递增/递减区间内可以判断，否则丢回去继续while

解法：

```
public int search(int[] A, int target) {
        // write your code here
        if(A.length == 0)
        return -1;

        int s = 0, e = A.length - 1;

        while(s + 1 < e){
            int m = s + (e - s)/2;
            if (A[m] == target) {
                return m;
            } 
            if(A[m] > A[s]){//m在上半段
                if(A[s] <= target && target <= A[m]){//在单调递增区间才能判定，否则还是2条断线，丢回去继续while
                    e = m;
                }else{
                    s = m;
                }
            }
            else{
                if(A[m] <= target && target <= A[e]){
                    s = m;
                }else{
                    e = m;
                }
            }
        }

        if(A[s] == target)
            return s;

        if(A[e] == target)
            return e;

            return -1;
    }
```