chapter-four/0800~0899/0845.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 }