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.
这道题可以和之前那个rotated sorted array的思路一样，cut haof of array every time来找target。
然后就没啥可说了，单独写一个find()method比较方便。
只是要注意这里的if的条件就要细致的多了，要多考虑，最好画个图（……我就画了，因为浆糊了不画图哈哈哈）
代码如下。~
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);
}
}