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

(i.e.,0 1 2 4 5 6 7might become4 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.

這道題可以和之前那個rotated sorted array的思路一樣，cut haof of array every time來找target。

然后就沒啥可說了，單獨寫一個find()method比較方便。

只是要注意這里的if的條件就要細致的多了，要多考慮，最好畫個圖（……我就畫了，因為漿糊了不畫圖哈哈哈）

代碼如下。~

public class Solution {    public int search(int[] nums, int target) {        int len=nums.length;        return find(nums,0,len-1,target);            }    public int find(int[] nums,int start,int end,int target){        if(start>end){            return -1;        }        int mid=(start+end)/2;        if(nums[mid]==target){            return mid;        }        if((target<nums[mid]&&target>=nums[start])||(target<nums[mid]&&nums[mid]<nums[start])||(target>=nums[start]&&nums[start]>nums[mid])){            return find(nums,start,mid-1,target);        }        return find(nums,mid+1,end,target);    }}