154 Find Minimum in Rotated Sorted Array 2

这道题与 153 Find Minimum in Rotated Sorted Array 基本相同，只是有可能有重复的数字，这样a[left] 就可能与a[mid]相同因此就多一个可能，因此在最坏的情况下复杂度会是O(n)

class Solution:# @param {integer[]} nums# @return {integer}def findMin(self, nums):l = len(nums) - 1if l <= 1:return min(nums)mid = l / 2if nums[0] < nums[mid]:return self.findMin([nums[0]] + nums[mid+1:])elif nums[0] > nums[mid]:return self.findMin(nums[1:mid+1])else:return min(self.findMin(nums[0:mid]),self.findMin(nums[mid+1:]))

转载于:https://www.cnblogs.com/dapanshe/p/4632808.html

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