0845. Longest Mountain in Array

845. Longest Mountain in Array #

题目 #

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?

题目大意 #

这道题考察的是滑动窗口的问题。

给出一个数组,要求求出这个数组里面“山”最长的长度。“山”的意思是,从一个数开始逐渐上升,到顶以后,逐渐下降。

解题思路 #

这道题解题思路也是滑动窗口,只不过在滑动的过程中多判断一个上升和下降的状态即可。

代码 #


package leetcode

func longestMountain(A []int) int {
	left, right, res, isAscending := 0, 0, 0, true
	for left < len(A) {
		if right+1 < len(A) && ((isAscending == true && A[right+1] > A[left] && A[right+1] > A[right]) || (right != left && A[right+1] < A[right])) {
			if A[right+1] < A[right] {
				isAscending = false
			}
			right++
		} else {
			if right != left && isAscending == false {
				res = max(res, right-left+1)
			}
			left++
			if right < left {
				right = left
			}
			if right == left {
				isAscending = true
			}
		}
	}
	return res
}


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