[Leetcode] (python): 081-search in rotated Sorted Array II

Source of the topic

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

Follow to "Search in rotated Sorted Array":
What if duplicates is allowed?

Would this affect the run-time complexity? How and why?

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

Test instructions Analysis

Input:

: Type Nums:list[int]
: Type Target:int

Output:

Rtype:bool

Conditions: An ordered array of flips, array elements may be repeated to determine whether the target is in the array

Topic ideas

The key is to distinguish the boundary, using first for the lower bound, last for the upper bound, mid as the middle point. Returns true if mid is target, otherwise, determines whether the mid,first,last is equal, if equal, narrows the search space, and then determines which interval the target is in, judging by: 1) the size of target and Nums[mid], Target and Nums The size of [first].

AC Code (PYTHON)

1 classsolution (object):2     defSearch (self, nums, target):3         """4 : Type Nums:list[int]5 : Type Target:int6 : Rtype:bool7         """8Size =Len (nums)9First =0TenLast = size-1 One          whileFirst <=Last : AMid = (last + first)/2 -             Print(First,mid, last) -             ifNums[mid] = =Target: the                 returnTrue -             ifNums[mid] = = Nums[first] = =Nums[last]: -First + = 1; Last-= 1 -             elifNums[first] <=Nums[mid]: +                 ifTarget < Nums[mid] andTarget >=Nums[first]: -Last = Mid-1 +                 Else: AFirst = mid + 1 at             Else: -                 ifTarget >= Nums[mid] andTarget <Nums[first]: -First = mid + 1 -                 Else: -Last = Mid-1 -  in  -  to         returnFalse +         

[Leetcode] (python): 081-search in rotated Sorted Array II