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

题目来源

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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.

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题意分析

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Input:

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

Output:

rtype: bool

Conditions:一个翻转的有序数组，数组元素可能重复，判断target是否在数组中

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题目思路

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关键就要区分边界，采用first表示下界，last表示上界，mid为中间点。如果mid为target，返回True；否则，判断mid，first，last是否相等，若相等则缩小搜索空间，之后判断target在哪个区间，判断条件为：1）target与nums[mid]的大小，target与nums[first]的大小。

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AC代码（Python）

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        size = len(nums)
        first = 0
        last = size - 1
        while first <= last:
            mid = (last + first) / 2
            print(first,mid, last)
            if nums[mid] == target:
                return True
            if nums[mid] == nums[first] == nums[last]:
                first += 1; last -= 1
            elif nums[first] <= nums[mid]:
                if target < nums[mid] and target >= nums[first]:
                    last = mid - 1
                else:
                    first = mid + 1
            else:
                if target >= nums[mid] and target < nums[first]:
                    first = mid + 1
                else:
                    last = mid - 1



        return False