33搜索旋转排序数组

题目：假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。你可以假设数组中不存在重复的元素。你的算法时间复杂度必须是 O(log n) 级别。
链接：https://leetcode-cn.com/problems/search-in-rotated-sorted-array

法一：自己的代码

思路：先判段[left,mid]是否为单增区间，如果是，判断target是否在该区间中，进而缩小搜索范围，如果不是，判断target是否在[mid,right]中，同理缩小搜索范围，注意用这个方法时，必须用left <= right判断，因为要用nums[right]判断与target的大小，所以right=len(nums)-1，用右闭区间，所以用left <= right，另外要注意[5,9],target=9这种情况。

# 原先的程序
# from typing import List
# class Solution:
#     def search(self, nums: List[int], target: int) -> int:
#         left = 0
#         right = len(nums) - 1
#         def recursion(left=0, right=right):
#             # if left == right:
#             #     return nums[left]
#             if nums[left] == target:
#                 return left
#             elif nums[right] == target:
#                 return right
#             elif left == right:
#                 return -1
#             mid = left + (right - left) // 2
#             # if target > nums[mid]:
#             if nums[mid] > nums[left]:
#                 if target > nums[mid] or target < nums[right]:
#                     return recursion(left=mid, right=right)
#                 else:
#                     return recursion(left=left, right=mid)
#             elif target > nums[left] or target < nums[mid]:
#                 return recursion(left=left, right=mid)
#             else:
#                 return recursion(left=mid, right=right)
#         return recursion(left, right)

# from typing import List
# class Solution:
#     def search(self, nums: List[int], target: int) -> int:
#         left = 0
#         right = len(nums) - 1
#         l = right
#         if right == -1:
#             return -1
#         def judge(left, right):
#             # 这里是
#             while left <= right:
#                 mid = (left + right) >> 1
#                 if nums[mid] == target:
#                     return mid
#                 if nums[mid] > target:
#                     right = mid - 1
#                 elif nums[mid] < target:
#                     left = mid + 1
#             if nums[left] == target:
#                 return left
#             else:
#                 return - 1
#         while left <= right:
#             mid = (left + right) >> 1
#             if nums[mid] == target:
#                 return mid
#             # 如果大于，说明左边为单增区间，
#             if nums[mid] > nums[left]:
#                 # 如果在该区间里面，直接搜索寻找
#                 if target >= nums[left] and target < nums[mid]:
#                     return judge(left, mid-1)
#                 # 否则说明在另一个区间里
#                 else:
#                     left = mid + 1
#             # 否则右边为单增区间
#             # mid+1有可能超过区间，所以取其和l的最小值
#             elif target >= nums[min(mid+1, l)] and target <= nums[right]:
#                 # mid+
#                 return judge(mid+1, right)
#             else:
#                 right = mid - 1
#         return left if nums[left] == target else -1

# 改进后的程序
from typing import List
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left = 0
        right = len(nums) - 1
        l = right
        if right == -1:
            return -1
        while left <= right:
            mid = (left + right) >> 1
            if nums[mid] == target:
                return mid
            # 如果大于，说明左边为单增区间，
            # 这里必须是等于等于，因为有[5,9],target=9这种情况
            if nums[mid] >= nums[left]:
                # 如果在该区间里面，直接搜索寻找
                if target >= nums[left] and target < nums[mid]:
                    right = mid - 1
                # 否则说明在另一个区间里
                else:
                    left = mid + 1
            # 否则右边为单增区间
            elif target > nums[mid] and target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
        return left if nums[left] == target else -1

if __name__ == '__main__':
    solution = Solution()
    # result = solution.search(nums=[6,7,8,9,10,11,12,1,2,3], target=4)
    result = solution.search(nums=[5,3], target=3)
    # result = solution.search(nums = [2,3,4,5,6,7,8,9,1], target = 3)
    print(result)

ttt