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1940. 排序数组之间的最长公共子序列 🔒

题目描述

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。

示例1:

输入: arrays = [[1,3,4],
               [1,4,7,9]]
输出: [1,4]
解释: 这两个数组中的最长子序列是[1,4]。

示例 2:

输入: arrays = [[2,3,6,8],
               [1,2,3,5,6,7,10],
               [2,3,4,6,9]]
输出: [2,3,6]
解释: 这三个数组中的最长子序列是[2,3,6]。

示例 3:

输入: arrays = [[1,2,3,4,5],
               [6,7,8]]
输出: []
解释: 这两个数组之间没有公共子序列。

 

限制条件:

  • 2 <= arrays.length <= 100
  • 1 <= arrays[i].length <= 100
  • 1 <= arrays[i][j] <= 100
  • arrays[i] 是严格递增排序.

解法

方法一

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class Solution:
    def longestCommomSubsequence(self, arrays: List[List[int]]) -> List[int]:
        n = len(arrays)
        counter = defaultdict(int)
        for array in arrays:
            for e in array:
                counter[e] += 1
        return [e for e, count in counter.items() if count == n]

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class Solution {
    public List<Integer> longestCommomSubsequence(int[][] arrays) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int[] array : arrays) {
            for (int e : array) {
                counter.put(e, counter.getOrDefault(e, 0) + 1);
            }
        }
        int n = arrays.length;
        List<Integer> res = new ArrayList<>();
        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
            if (entry.getValue() == n) {
                res.add(entry.getKey());
            }
        }
        return res;
    }
}

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class Solution {
public:
    vector<int> longestCommomSubsequence(vector<vector<int>>& arrays) {
        unordered_map<int, int> counter;
        vector<int> res;
        int n = arrays.size();
        for (auto array : arrays) {
            for (auto e : array) {
                counter[e] += 1;
                if (counter[e] == n) {
                    res.push_back(e);
                }
            }
        }
        return res;
    }
};

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func longestCommomSubsequence(arrays [][]int) []int {
    counter := make(map[int]int)
    n := len(arrays)
    var res []int
    for _, array := range arrays {
        for _, e := range array {
            counter[e]++
            if counter[e] == n {
                res = append(res, e)
            }
        }
    }
    return res
}

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/**
 * @param {number[][]} arrays
 * @return {number[]}
 */
var longestCommonSubsequence = function (arrays) {
    const m = new Map();
    const rs = [];
    const len = arrays.length;
    for (let i = 0; i < len; i++) {
        for (let j = 0; j < arrays[i].length; j++) {
            m.set(arrays[i][j], (m.get(arrays[i][j]) || 0) + 1);
            if (m.get(arrays[i][j]) === len) rs.push(arrays[i][j]);
        }
    }
    return rs;
};

方法二

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class Solution:
    def longestCommomSubsequence(self, arrays: List[List[int]]) -> List[int]:
        def common(l1, l2):
            i, j, n1, n2 = 0, 0, len(l1), len(l2)
            res = []
            while i < n1 and j < n2:
                if l1[i] == l2[j]:
                    res.append(l1[i])
                    i += 1
                    j += 1
                elif l1[i] > l2[j]:
                    j += 1
                else:
                    i += 1
            return res

        n = len(arrays)
        for i in range(1, n):
            arrays[i] = common(arrays[i - 1], arrays[i])
        return arrays[n - 1]

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