845. Longest Mountain in Array

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

B.length >= 3
There exists some 0 < i < B.length - 1 such that B[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 array A of integers, return the length of the longest mountain.

Return 0 if 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:

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

Follow up:

Can you solve it using only one pass?
Can you solve it in O(1) space?

class Solution:
    def longestMountain(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        length = 0
        for i in range(len(A)):
            temp = 0
            if i>0 and i<len(A)-1 and A[i]>A[i-1] and A[i]>A[i+1]:
                temp = 1
                start,end = i,i
                while start>0 and A[start]>A[start-1]:
                    start -= 1
                    temp += 1
                while end<len(A)-1 and A[end]>A[end+1]:
                    end += 1
                    temp +=1
            length = max(temp,length)
        return length

从中间到两边扩展，如果考虑起点和终点比较复杂，考虑山顶更容易。