字符串算法——旋轉數組中查找目標值(Search in Rotated Sorted Array)

問題：
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.
解決思路：如果不考慮複雜度，可以對數組每個元素進行比較，看是否存在查找的目標值，如果考慮複雜度，可以利用數組原來有序的先驗知識，必然數組一邊是有序的，一邊是無序的，使用二分查找的方法。

class Solution {
    public int search(int[] nums, int target) {
        int len = nums.length;
        if(len==0)return -1;
        return dfs(nums,target,0,len-1);


    }
    private int dfs(int []nums,int target,int min,int max){
        if(min>max){
            return -1;
        }
        int mid = (min+max)/2;//中間索引
        if(nums[mid] ==target)return mid;
        if(nums[mid]>nums[min]){//中間索引左側是有序的
            if(nums[mid]>target && nums[min]<target){//落在索引左側
                return dfs(nums,target,min+1,mid-1);
            }
            if(nums[min] == target){
                return min;
            }
            return dfs(nums,target,mid+1,max);//落在索引右側
        }else {//索引右側是有序的
            if(nums[mid]<target && nums[max]>target){//落在索引右側
                return dfs(nums,target,mid+1,max-1);
            }
            if(nums[max] ==target){
                return max;
            }
            return dfs(nums,target,min,mid-1);//落在索引左側
        }
    }
}