Description: Suppose an array of length n
sorted in ascending order is rotated between 1
and n
times. For example, the array nums = [0,1,4,4,5,6,7]
might become:
[4,5,6,7,0,1,4]
if it was rotated4
times.[0,1,4,4,5,6,7]
if it was rotated7
times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]]
1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]]
.
Given the sorted rotated array nums
that may contain duplicates, return the minimum element of this array.
Link: 154. Find Minimum in Rotated Sorted Array II
Examples:
Example 1: Input: nums = [1,3,5] Output: 1 Example 2: Input: nums = [2,2,2,0,1] Output: 0 Example 3: Input: nums = [1,3,3] Output: 1 Example 4: Input: nums = [3,1,3,3] Output: 1
思路: 不同于153,这个题目加了重复的数字,使用原来一样的code,example3过不了,nums[mid] == nums[r] 的情况,r -= 1,先前移动,退化为顺序查找。复杂度O(n).
class Solution(object): def findMin(self, nums): """ :type nums: List[int] :rtype: int """ n = len(nums) l, r = 0, n-1 while l < r: mid = int((l+r)/2) if nums[mid] < nums[r]: r = mid elif nums[mid] == nums[r]: r -= 1 else: l = mid+1 return nums[l]
Reference: https://www.cnblogs.com/grandyang/p/4040438.html
日期: 2021-04-11