leetcode【845】Longest Mountain in Array

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最新推荐文章于 2025-05-31 00:45:00 发布

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本文详细解析LeetCode题目845，寻找数组中最长的山脉子数组，即连续子数组中元素先递增再递减的情况。通过遍历数组并比较相邻元素，统计最长山脉子数组的长度。

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写在最前面：不得不说，python对于数据结构的处理非常自然，几乎就是你怎么想的直接写出来就可以，不用太过在意转化，代价就是，性能较差

leetcode【845】Longest Mounta 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

简而言之就是对一个数组，找出长度最大的类似山峰的子数组

对于这道题的思路就是，比较A[i]两边的大小，向左，向右，然后分别统计两侧的值，加起来

class Solution:
    def longestMountain(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        length = len(A)
        top = 0
        for i in range(1, length):
            xleft = 0
            xright = 0
            left = i - 1
            right = i + 1
            k = A[i]
            while left >= 0 and A[left] < k:
                k = A[left]
                left -= 1
                xleft += 1
            k = A[i]
            while right <= length - 1 and A[right] < k:
                k = A[right]
                right += 1
                xright += 1

            if xleft >= 1 and xright >= 1:
                top = max(top, xright + xleft + 1)
        return top

怎么说呢，leetcode没过，因为超时了，超时的用例实在有点恐怖，相同的java写的是过了的，有时间会想想怎么优化好一点

抛砖引玉吧，如果有更好的思路欢迎私我~