題目
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.
簡單的做法
????順序掃描一遍
代碼
class Solution:
# @param {integer[]} nums
# @return {integer}
def findMin(self, nums):
i = 0
while i+1 < len(nums) and nums[i+1] > nums[i]:
i += 1
if i == len(nums)-1:
return nums[0]
return nums[i+1]
二分查找的做法
當nums[m] == nums[l]，此時或者l==r或者r=l+1，直接比較即可；當nums[m] > nums[l]時，左邊的最小元素是nums[l]，右邊可能有比它小的也可能沒有，找出右邊最小的與nums[l]比較即可；當nums[m] < nums[l]時，最小元素一定出現在[l,m]中，需包含m。
代碼
class Solution(object):
def search(self, nums, l, r):
m = l + (r-l)/2
if nums[m] == nums[l]:
return min(nums[l],nums[r])
elif nums[m] > nums[l]:
return min(nums[l],self.search(nums, m+1, r))
else:
return self.search(nums, l, m)
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if nums==None or len(nums)==0:
return None
return self.search(nums,0,len(nums)-1)
