README_EN.md

852. Peak Index in a Mountain Array

中文文档

Description

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 <= 104
• 0 <= arr[i] <= 106
• arr is guaranteed to be a mountain array.

 

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

Solutions

Binary search.

Python3

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        n = len(arr)
        left, right = 1, n - 2
        while left < right:
            mid = (left + right) // 2
            if arr[mid] < arr[mid + 1]:
                left = mid + 1
            else:
                right = mid
        return right

Java

class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int n = arr.length;
        int left = 1, right = n - 2;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] < arr[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return right;
    }
}

Go

func peakIndexInMountainArray(arr []int) int {
	n := len(arr)
	left, right := 1, n-2
	for left < right {
		mid := left + (right-left)/2
		if arr[mid] < arr[mid+1] {
			left = mid + 1
		} else {
			right = mid
		}
	}
	return right
}

C++

class Solution {
public:
    int peakIndexInMountainArray(vector<int>& arr) {
        int left = 1, right = arr.size() - 2;
        while (left < right) {
            int mid = left + right >> 1;
            if (arr[mid] < arr[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return right;
    }
};

JavaScript

/**
 * @param {number[]} arr
 * @return {number}
 */
var peakIndexInMountainArray = function(arr) {
    let left = 1;
    let right = arr.length - 2;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (arr[mid] < arr[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
};

...