Question

941. Valid Mountain Array

中文文档

Description

Given an array of integers arr, return _true if and only if it is a valid mountain array_.

Recall that arr is a mountain array if and only if:

• 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]

 

Example 1:

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

Example 2:

Input: arr = [3,5,5]
Output: false

Example 3:

Input: arr = [0,3,2,1]
Output: true

 

Constraints:

• 1 <= arr.length <= 104
• 0 <= arr[i] <= 104

Solutions

Solution 1

class Solution:
  def validMountainArray(self, arr: List[int]) -> bool:
    n = len(arr)
    if n < 3:
      return False
    l, r = 0, n - 1
    while l + 1 < n - 1 and arr[l] < arr[l + 1]:
      l += 1
    while r - 1 > 0 and arr[r] < arr[r - 1]:
      r -= 1
    return l == r

class Solution {

  public boolean validMountainArray(int[] arr) {
    int n = arr.length;
    if (n < 3) {
      return false;
    }
    int l = 0, r = n - 1;
    while (l + 1 < n - 1 && arr[l] < arr[l + 1]) {
      ++l;
    }
    while (r - 1 > 0 && arr[r] < arr[r - 1]) {
      --r;
    }
    return l == r;
  }
}

class Solution {
public:
  bool validMountainArray(vector<int>& arr) {
    int n = arr.size();
    if (n < 3) return 0;
    int l = 0, r = n - 1;
    while (l + 1 < n - 1 && arr[l] < arr[l + 1]) ++l;
    while (r - 1 > 0 && arr[r] < arr[r - 1]) --r;
    return l == r;
  }
};

func validMountainArray(arr []int) bool {
  n := len(arr)
  if n < 3 {
    return false
  }
  l, r := 0, n-1
  for l+1 < n-1 && arr[l] < arr[l+1] {
    l++
  }
  for r-1 > 0 && arr[r] < arr[r-1] {
    r--
  }
  return l == r
}