LeetCode 845 Longest Mountain in Array (模拟)

最新推荐文章于 2025-05-31 00:45:00 发布

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最新推荐文章于 2025-05-31 00:45:00 发布

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分类专栏： Leetcode Medium LeetCode 文章标签：模拟

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本文介绍了一个高效的算法，用于解决LeetCode上的最长山脉子数组问题。通过一次遍历和O(1)空间复杂度，该算法能够找到给定整数数组中最长的山脉子数组的长度。文章详细解释了算法的工作原理，包括如何使用incr和decr变量来跟踪连续序列的递增和递减部分。

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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?

题目链接：https://leetcode.com/problems/longest-mountain-in-array/

题目分析：要求一次遍历且O(1)空间，需要分一些情况讨论

设变量incr和decr分别表示到当前位置pos时之前连续序列的递增数和递减数，若A[pos]<=A[pos+1]，此时无论decr的值是多少都需要清0，但若incr不为0则说明之前存在一个mountain，更新答案，对incr也清0（此时的incr表示之前一组连续上升序列的长度）若A[pos]==A[pos+1]，则incr=0，若A[pos]<A[pos+1]，incr加1，这都很好理解，递减的情况只需要更新decr即可

击败100%

class Solution {
    public int longestMountain(int[] A) {
        int incr = 0, decr = 0, ans = 0;
        for (int i = 0; i < A.length - 1; i++) {
            if (A[i] <= A[i + 1]) {
                if (decr > 0) {
                    if (incr > 0) {
                        ans = Math.max(ans, decr + incr + 1);
                        incr = 0;
                    }
                    decr = 0;
                }
                if (A[i] < A[i + 1]) {
                    incr++;
                } else {
                    incr = 0;
                }
            } else if (A[i] > A[i + 1]) {
                decr++;
            }
        }
        if (decr > 0 && incr > 0) {
            ans = Math.max(ans, decr + incr + 1);
        }
        return ans;
    }
}