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;
    }
}