# 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 < arr < ... < 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`
• 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
}