845. Longest Mountain in Array

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本文解析了如何使用双指针和动态规划解决LeetCode上的‘最长山峰数组’问题，介绍了两种不同的算法实现，包括一次遍历和空间优化版本。

You may recall that an array arr is a mountain array if and only if:

arr.length >= 3
There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Given an integer array arr, return the length of the longest subarray, which is a mountain. Return 0 if there is no mountain subarray.

Example 1:

Input: arr = [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: arr = [2,2,2]
Output: 0
Explanation: There is no mountain.

Constraints:

1 <= arr.length <= 104
0 <= arr[i] <= 104

Follow up:

Can you solve it using only one pass?
Can you solve it in O(1) space?

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/longest-mountain-in-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

用一个变量表示当前走向是上、下或者相等，初始值是相等。每次都计算出最长子串。

class Solution {
    public int longestMountain(int[] arr) {
        if (arr.length <= 2) {
            return 0;
        }
        int index = 1;
        int num = 0;
        int up = 0;
        int max = 0;
        while (index < arr.length) {
            if (arr[index] > arr[index - 1]) {
                if (up == 1) {
                    num++;
                } else if (up == -1) {
                    up = 1;
                    num = 2;
                } else {
                    num = 2;
                    up = 1;
                }
            } else if (arr[index] < arr[index - 1]) {
                if (up == 1) {
                    num++;
                    up = -1;
                } else if (up == -1) {
                    num++;
                } else {
                    num = 0;
                    up = 0;
                }
            } else if (arr[index] == arr[index - 1]) {
                num = 0;
                up = 0;
            }
            index++;
            if (up == -1) {
                // 只要向下走的时候，才需要计算结果
                max = Math.max(max, num);
            }
        }
        return max;
    }
}

方法二：

两个dp数组，分别用两个 dp 数组 up 和 down，其中 up[i] 表示以 i 位置为终点的最长递增数列的个数，down[i] 表示以 i 位置为起点的最长递减数列的个数。则结果就是up[i] + down[i] + 1

class Solution {
    public int longestMountain(int[] arr) {
        int res = 0, n = arr.length;
        int []up = new int[arr.length];
        int []down = new int[arr.length];
        
        for (int i = n - 2; i >= 0; --i) {
            if (arr[i] > arr[i + 1]) down[i] = down[i + 1] + 1;
        }
        for (int i = 1; i < n; ++i) {
            if (arr[i] > arr[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;
    }
}