845.Longest-Mountain-in-Array.md

845. Longest Mountain in Array

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题目地址

https://leetcode.com/problems/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

Follow up:
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;
  }