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# 845.Longest-Mountain-in-Array

## 题目描述

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

## 代码

### Approach #1 Two Pointer

Time: O(N) && Space: O(1)
class Solution {
public int longestMountain(int[] A) {
int N = A.length;
int ans = 0;
int base = 0;
while (base < N) {
int end = base;
if (end+1 < N && A[end] < A[end+1]) {
while (end+1 < N && A[end] < A[end+1]) {
end++;
}
if (end+1 < N && A[end] > A[end+1]) {
while (end+1 < N && A[end] > A[end+1]) {
end++;
}
ans = Math.max(ans, end - base + 1);
}
}
base = Math.max(end, base + 1);
}
return ans;
}
}

### Approach #2

public int longestMountain(int[] A) {
int N = A.length;
int res = 0;
int[] up = new int[N];
int[] down = new int[N];
for (int i = N - 2; i >= 0; --i) {
if (A[i] > A[i + 1]) down[i] = down[i + 1] + 1;
}
for (int i = 0; i < N; ++i) {
if (i > 0 && A[i] > A[i - 1]) up[i] = up[i - 1] + 1;
if (up[i] > 0 && down[i] > 0) res = Math.max(res, up[i] + down[i] + 1);
}
return res;
}