LeetCode 33.Search in Rotated Sorted Array（M）

题目：

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.

思路：

1.当数组长度为0时，return -1；

2.当数组长度为1时，比较nums[0]与target关系;

3.当数组长度大于1，且为rotated sorted array时，根据nums[mid]与nums[start]关系判断如何移动；

4.当数组长度大于1，但不是rotated sorted array时，用传统二分法；

5.注意，不仅仅要比较target与nums[mid]关系，还要比较其与nums[start]和nums[end]关系，否则会出现死循环。

代码：

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