[LeetCode] 1940. Longest Common Subsequence Between Sorted Arrays

Given an array of integer arrays arrays where each arrays[i] is sorted in strictly increasing order, return an integer array representing the longest common subsequence between all the arrays.

A subsequence is a sequence that can be derived from another sequence by deleting some elements (possibly none) without changing the order of the remaining elements.

Example 1:

Input: arrays = [[1,3,4],
                 [1,4,7,9]]
Output: [1,4]
Explanation: The longest common subsequence in the two arrays is [1,4].

Example 2:

Input: arrays = [[2,3,6,8],
                 [1,2,3,5,6,7,10],
                 [2,3,4,6,9]]
Output: [2,3,6]
Explanation: The longest common subsequence in all three arrays is [2,3,6].

Example 3:

Input: arrays = [[1,2,3,4,5],
                 [6,7,8]]
Output: []
Explanation: There is no common subsequence between the two arrays.

Constraints:

• 2 <= arrays.length <= 100
• 1 <= arrays[i].length <= 100
• 1 <= arrays[i][j] <= 100
• arrays[i] is sorted in strictly increasing order.

排序数组之间的最长公共子序列。

给定一个由整数数组组成的数组 arrays，其中 arrays[i] 是严格递增排序的，返回一个表示所有数组之间的最长公共子序列的整数数组。

子序列是从另一个序列派生出来的序列，删除一些元素或不删除任何元素，而不改变其余元素的顺序。

这道题我给出一个类似 counting sort 的思路。同时这道题跟最长公共子序列这个LC常考的考点没什么太大关系。

注意题目给的二维数组并不是一个完美的矩阵，每一行的元素个数并不是一样的。同时注意题目给定的元素区间是 [1, 100]，所以我可以创建一个长度为 101 的数组 map，之后开始遍历一遍这个二维数组，统计一下每个数字都出现了几次。因为题目问的是统计所有数组中出现的最长公共子序列，而且每个子数组又保持递增，所以当我们统计完毕之后，去 map 里找出现次数 == 子数组个数的数字，即是题目要求的数字。

时间O(n^2)

空间O(n)

Java实现

class Solution {
    public List<Integer> longestCommonSubsequence(int[][] arrays) {
        int[] map = new int[101];
        for (int[] row : arrays) {
            for (int n : row) {
                map[n]++;
            }
        }
        
        List<Integer> res = new ArrayList<>();
        for (int number = 1; number <= 100; number++) {
            if (map[number] == arrays.length) {
                res.add(number);
            }
        }
        return res;
    }
}

 

LeetCode 题目总结