Leetcode 162. Find Peak Element
https://leetcode.com/problems/find-peak-element/
Medium
A peak element is an element that is greater than its neighbors.
Given an input array nums
, where nums[i] ≠ nums[i+1]
, find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[-1] = nums[n] = -∞
.
Example 1:
Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.
Example 2:
Input: nums = [
1,2,1,3,5,6,4]
Output: 1 or 5
Explanation: Your function can return either index number 1 where the peak element is 2,
or index number 5 where the peak element is 6.
Note:
Your solution should be in logarithmic complexity.
- 二分查找。学习下思路分析,搞清楚为什么二分查找可以。
- https://leetcode.com/problems/find-peak-element/solution/
- Python3应该用//,懒得改了。 E.g. int (3 / 2) == 3 // 2
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution: 2 def findPeakElement1(self, nums: List[int]) -> int: 3 left, right = 0, len(nums) - 1 4 5 while left < right: 6 middle = int( (left + right) / 2) 7 8 if nums[middle] > nums[middle + 1]: 9 right = middle 10 else: 11 left = middle + 1 12 13 return left 14 15 def findPeakElement(self, nums: List[int]) -> int: 16 if not nums: 17 return 0 18 19 def helper(nums, left, right): 20 if left == right: 21 return left 22 23 middle = int( (left + right) / 2 ) 24 25 if nums[middle] > nums[middle + 1]: 26 return helper(nums, left, middle) 27 else: 28 return helper(nums, middle + 1, right) 29 30 return helper(nums, 0, len(nums) - 1)