[LeetCode]題解（python）：081 - Search in Rotated Sorted Array II

標籤：

題目來源

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

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

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

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

題意分析

Input:

:type nums: List[int]
:type target: int

Output:

rtype: bool

Conditions:一個翻轉的有序數組，數組元素可能重複，判斷target是否在數組中

題目思路

關鍵就要區分邊界，採用first表示下界，last表示上界，mid為中間點。如果mid為target，返回True；否則，判斷mid，first，last是否相等，若相等則縮小搜尋空間，之後判斷target在哪個區間，判斷條件為：1）target與nums[mid]的大小，target與nums[first]的大小。

AC代碼（Python）

 1 class Solution(object): 2     def search(self, nums, target): 3         """ 4         :type nums: List[int] 5         :type target: int 6         :rtype: bool 7         """ 8         size = len(nums) 9         first = 010         last = size - 111         while first <= last:12             mid = (last + first) / 213             print(first,mid, last)14             if nums[mid] == target:15                 return True16             if nums[mid] == nums[first] == nums[last]:17                 first += 1; last -= 118             elif nums[first] <= nums[mid]:19                 if target < nums[mid] and target >= nums[first]:20                     last = mid - 121                 else:22                     first = mid + 123             else:24                 if target >= nums[mid] and target < nums[first]:25                     first = mid + 126                 else:27                     last = mid - 128 29 30 31         return False32         

 

[LeetCode]題解（python）：081 - Search in Rotated Sorted Array II