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]

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