http://www.lintcode.com/en/problem/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 For [4, 5, 1, 2, 3] and target=1, return 2. For [4, 5, 1, 2, 3] and target=0, return -1.

解题：反转数组里面找目标值，和上一题一样需要判断取中间值的时候左半边是sorted还是右半边是sorted。如果右边是sorted，我们可以看target在不在mid和end之间，如果在走右边，也就是start ＝ mid， 否则走左边。左边sorted同理。

public class Solution {
    /** 
     *@param A : an integer rotated sorted array
     *@param target :  an integer to be searched
     *return : an integer
     */
    public int search(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0) {
            return -1;
        }
        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return mid;
            }

            //need to check which side is sorted
            //right side is sorted
            if (A[mid] <= A[start]) {
                //if target is on sorted right side
                if (A[mid] < target && target <= A[end]) {
                    start = mid;
                } else {
                //if target is not on right side, go left
                    end = mid;
                }
            } else {
                //left side is sorted
                //if the target is on sorted left side
                if (A[start] <= target && target < A[mid]) {
                    end = mid;
                } else {
                //if target is not on left side, go right
                    start = mid;
                }
            }
        }
        if (A[start] == target) {
            return start;
        }
        if (A[end] == target) {
            return end;
        }
        return -1;
    }
}