Let's call any (contiguous) subarray B (of A)amountainif the following properties hold:

• B.length >= 3
• There exists some0 < i< B.length - 1such thatB[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]

(Note that B could be any subarray of A, including the entire array A.)

Given an arrayAof integers,return the length of the longestmountain.

Return0if there is no mountain.

Example 1:

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

Note:

1. 0 <= A.length <= 10000
2. 0 <= A[i] <= 10000

Follow up:

• Can you solve it using only one pass?
• Can you solve it inO(1)space?

class Solution {
    public int longestMountain(int[] A) {
            int n=A.length;
            if(n<3)return 0;

            int left=0;int right;int max=0;

            while(left<n-2){
                
 
//skip decending and equal array
                while(left<n-1 && A[left]>=A[left+1]){
                    left++;
                }
                right=left+1;
                //mountain up
                while(right<n-1 && A[right]<A[right+1]){
                    right++;
                }
                //mountain down
                while(right<n-1 && A[right]>A[right+1]){
                    right++;
                    //update the max value
                    max=Math.max(max,right-left+1);
                }
                left=right;
            }
            return max;
        }
}

好耶