题目地址：https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/description/
题目描述：
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
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).
Find the minimum element.
Thearray may contain duplicates.
题目大意
找出可能含有重复数字的旋转有序数组中的最小值。
解题方法
这个题就是剑指offer中的原题。
思想就是如果出现了重复数字，那么二分查找就没有作用了，必须使用顺序查找了。
class Solution(object):
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
p1, p2 = 0, len(nums) - 1
mid = p1
while nums[p1] >= nums[p2]:
if p2 - p1 == 1:
mid = p2
break
mid = (p1 + p2) / 2
if nums[mid] == nums[p1] and nums[mid] == nums[p2]:
return self.minInOrder(nums, p1, p2)
if nums[mid] >= nums[p1]:
p1 = mid
elif nums[mid] <= nums[p2]:
p2 = mid
return nums[mid]
def minInOrder(self, nums, index1, index2):
n1 = nums[index1]
for i in range(index1 + 1, index2):
if n1 > nums[i]:
return nums[i]
return n1
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