852. Peak Index in a Mountain Array-山脉数组的峰顶索引

Let’s call an array arr a mountain if the following properties hold:

• arr.length >= 3
• There exists some i with 0 < i < arr.length – 1 such that:
  • arr[0] < arr[1] < … arr[i-1] < arr[i]
  • arr[i] > arr[i+1] > … > arr[arr.length – 1]

Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < … arr[i – 1] < arr[i] > arr[i + 1] > … > arr[arr.length – 1].

Example 1:
Input: arr = [0,1,0]
Output: 1

Example 2:
Input: arr = [0,2,1,0]
Output: 1

Example 3:
Input: arr = [0,10,5,2]
Output: 1

Example 4:
Input: arr = [3,4,5,1]
Output: 2

Example 5:
Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2

Constraints:

• 3 <= arr.length <= 10^4
• 0 <= arr[i] <= 10^6
• arr is guaranteed to be a mountain array.

Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?

Solution in python:

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        i = 0
        while arr[i] <= arr[i+1]:
            i += 1
        return i