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;
    }
}