LeetCode 81.Search in Rotated Sorted Array II(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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

思路：

1.当数组长度为0时，return false;

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，见下图：

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

代码：

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