Description

An array arr a mountain if the following properties hold:

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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 a mountain array arr, return the index i such that arr[0] < arr[1] < ... < arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

You must solve it in O(log(arr.length)) time complexity.

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:

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3 <= arr.length <= 10^5
0 <= arr[i] <= 10^6
arr is guaranteed to be a mountain array.

Solution

Use binary search to solve this problem. For any middle index, it either locates at the left of the peak, or the right of the peak. If at the left, then discard the left half, otherwise discard the right half.

Time complexity: o ( log ⁡ n ) o(\log n) o(logn)

Space complexity: o ( 1 ) o(1) o(1)

Code

python3 复制代码

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        left, right = 0, len(arr) - 1
        while left < right:
            mid = (left + right) >> 1
            if arr[mid - 1] < arr[mid] < arr[mid + 1]:
                left = mid + 1
            elif arr[mid - 1] > arr[mid] > arr[mid + 1]:
                right = mid
            else:
                return mid