845. Longest Mountain in Array

Medium

You may recall that an array arr is a mountain array if and only if:

• arr.length >= 3

• There exists some index i (0-indexed) 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, return the length of the longest subarray, which is a mountain. Return 0 if there is no mountain subarray.

Example 1:

Input: arr = [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.

Example 2:

Input: arr = [2,2,2]
Output: 0
Explanation: There is no mountain.

Constraints:

• 1 <= arr.length <= 10^4

• 0 <= arr[i] <= 10^4

Follow up:

• Can you solve it using only one pass?

• Can you solve it in O(1) space?

解題

func longestMountain(arr []int) int {
    ans := 0
    for i:=1; i<len(arr) - 1; i++ {
        l := findMountain(arr, i)
        if l > ans {
            ans = l
        } 
    }

    return ans
}

func findMountain(arr []int, index int) int {
    left := 0
    for i:=index - 1; i >= 0; i-- {
        if arr[i] >= arr[i + 1] {
            break
        }
        left++
    }

    if left == 0 { return 0 }

    right := 0
    for i:=index + 1; i < len(arr); i++ {
        if arr[i] >= arr[i - 1] {
            break
        }
        right++
    }

    if right == 0 { return 0 }
    return left + right + 1
}