Level：

  Medium

题目描述：

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

思路分析：

  有序旋转数组，设置两个指针，left和right，分别指向数组的左端和右端，求数组的mid，如果mid的值大于left的值，那么旋转点在mid后面，如果小于left，则证明旋转点在mid前面。

  如果旋点在mid的后面，并且target的值大于left处的值，小于mid处的值，那么接下来就可以在left到mid之间进行二分查找target，否则对mid+1到right这部分数组进行递归操作。

  如果旋点在mid的前面，并且target的值大于mid处的值，小于right处的值，那么接下来就可以在mid到right之间进行二分查找target，否则对left到mid-1这部分数组进行递归操作。

代码：

public class Solution{public int search(int []nums,int target){if(nums==null||nums.length==0)return -1;int res=find(nums,0,nums.length-1,target);return res;}public int find(int []nums,int left,int right,int target){if(nums[left]==target)return left;if(nums[right]==target)return right;int mid=(left+right)/2;if(nums[mid]==target)return mid;if(left>right)return -1;if(nums[mid]>nums[left]){ //证明旋转点在mid后面if(target>nums[left]&&target<nums[mid]){return binarySearch(nums,left,mid-1,target);}else{return find(nums,mid+1,right,target);}}if(nums[mid]<nums[left]){ //证明旋转点在mid的前面if(target>nums[mid]&&target<nums[right]){return binarySearch(nums,mid+1,right,target);}else{return find(nums,left,mid-1,target);}}return -1;}public int binarySearch(int []nums,int left,int right,int target){if(left<=right){int mid=(left+right)/2;if(nums[mid]==target)return mid;if(target>nums[mid]){return binarySearch(nums,mid+1,right,target);}if(target<nums[mid]){return binarySearch(nums,left,mid-1,target);}}return -1;}
}