【LeetCode每天一题】Search in Rotated Sorted Array（在旋转数组中搜索）

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.Your algorithm's runtime complexity must be in the order of O(log n).

Example 1: Input: nums = [4,5,6,7,0,1,2] ,         target = 0                         Output: 4

Example 2: Input: nums =[4,5,6,7,0,1,2]           , target = 3                          Output: -1

思路

---

　　这道题的主要思路我们可以利用二分查找的办法来解决问题，只不过相对于普遍性二分查找，旋转数组的情况会稍微复杂一些，主要是在缩小边界的判断上。如果nums[middle] 大于nums[start]时，我们还需要更具target和nums[start]、nums[middle]的大小关系来决定移动左边的指针还是移动右边的指针。
　　时间夫复杂度为O(nlog n),空间复杂度为O(1)。
图示

---

解决代码

---

 class Solution:
     def search(self, nums, target):
         """
         :type nums: List[int]
         :type target: int
         :rtype: int
         """
         if not nums:
             return -1
         start, end = 0, len(nums)-1
         while start <= end:             # 循环条件
             middle = start+((end-start)>>1)      # 求中间值
             if nums[middle] == target:       # 找到直接返回下标
                 return middle
             elif nums[start] <= nums[middle]:     # 先判断nums[start] 和nums[middle]的关系
                 if nums[start] <= target and target < nums[middle]:  # 在判断start和middle对应元素与target的关系，决定相应指针移动方向
                     end = middle -1
                 else:
                     start = middle + 1
             else:
                 if nums[middle] < target and target <= nums[end]:     # 判断之间的关系middle和end对应元素yutarget之间的关系。
                     start = middle + 1
                 else:
                     end = middle -1
         return -1           # 没有找到，返回-1