Leetcode Find Minimum in Rotated Sorted Array I and II

Suppose a sorted array 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.

You may assume no duplicate exists in the array.

在任何一个sublist中，如果头元素大于尾元素，那么这个minimum一定在这个sublist中间的某一个位置。可以用二分法找到这个元素，复杂度是O(logN)。本题O(N)的算法也可以通过OJ，思路就是最简答的从头元素往后比，直到出现 nums[i] > nums[i+1]的情况，那么nums[i+1]就是Minimum element。

 class Solution(object):
     def findMin(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         n = len(nums)
         left = 0
         right = n-1
         while left < right:
             mid = (left+right)//2
             if nums[mid] > nums[right]:
                 left = mid + 1
             else:
                 right = mid
         return nums[left]
         

以下引用来自： http://bangbingsyb.blogspot.com/2014/11/leecode-find-minimum-in-rotated-sorted.html

和Search in Rotated Sorted Array II这题换汤不换药。同样当A[mid] = A[end]时，无法判断min究竟在左边还是右边。

 

3 1 2 3 3 3 3 
3 3 3 3 1 2 3

 

但可以肯定的是可以排除A[end]：因为即使min = A[end]，由于A[end] = A[mid]，排除A[end]并没有让min丢失。所以增加的条件是：

 

A[mid] = A[end]：搜索A[start : end-1]

 

 class Solution(object):
     def findMin(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         n = len(nums)
         left = 0
         right = n-1
         while left < right:
             mid = (left+right)//2
             if nums[mid] > nums[right]:
                 left = mid + 1
             elif nums[mid] < nums[right]:
                 right = mid
             else:
                 right -= 1

         return nums[left]