【Day 57 】2024-01-11 - 1143.最长公共子序列

[azl397985856]

1143.最长公共子序列

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/longest-common-subsequence

前置知识

• 数组
• 动态规划

  题目描述

  给定两个字符串  text1 和  text2，返回这两个字符串的最长公共子序列的长度。

一个字符串的   子序列   是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。 例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的「公共子序列」是这两个字符串所共同拥有的子序列。

若这两个字符串没有公共子序列，则返回 0。

示例 1:

输入：text1 = "abcde", text2 = "ace" 输出：3
解释：最长公共子序列是 "ace"，它的长度为 3。 示例 2:

输入：text1 = "abc", text2 = "abc" 输出：3 解释：最长公共子序列是 "abc"，它的长度为 3。 示例 3:

输入：text1 = "abc", text2 = "def" 输出：0 解释：两个字符串没有公共子序列，返回 0。

提示:

1 <= text1.length <= 1000 1 <= text2.length <= 1000 输入的字符串只含有小写英文字符。

[Danielyan86]

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        # Get the lengths of the input texts
        m, n = len(text1), len(text2)

        # Create a 2D array to store the length of the common subsequence
        # dp[i][j] represents the length of the common subsequence of text1[:i] and text2[:j]
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        # Iterate through the texts to fill the dp array
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                # If the current characters match, extend the common subsequence
                if text1[i - 1] == text2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                # If the characters don't match, take the maximum length from the previous rows or columns
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        # The final value in the dp array represents the length of the longest common subsequence
        return dp[m][n]

[larscheng]

思路

• dp[i][j] 代表text1前i个字符text1[0:i-1]，与text2前j个字符text2[0:j-1] 的最长公共子序列长度
• 当text[i-1]==text[j-1]时,说明两个字符串最后一个字符相同
  • 此时在text1前i-1个字符，与text2前j-1个字符的最长公共子序列长度(dp[i-1][j-1])的基础上+1即可
  • dp[i][j] = dp[i-1][j-1] + 1
• 当text[i-1]!=text[j-1]时,说明两个字符串最后一个字符不同
  • 此时需要考虑两种情况
    • text1前i-1个字符 与 text2前j个字符的 最大公共子序列长度 dp[i-1][j]
    • text1前i个字符 与 text2前j-1个字符的 最大公共子序列长度 dp[i][j-1]
  • 取两者最大值即可 dp[i][j] = Max(dp[i-1][j], dp[i][j-1])

    代码

    class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
    int m = text1.length(), n = text2.length();
    int[][] dp = new int[m + 1][n + 1];
    for (int i = 1; i <= m; i++) {
        char c1 = text1.charAt(i - 1);
        for (int j = 1; j <= n; j++) {
            char c2 = text2.charAt(j - 1);
            if (c1 == c2) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    return dp[m][n];
    }
    }

    复杂度

• 时间复杂度：O(m*n)
• 空间复杂度：O(m*n)

[adfvcdxv]

var longestCommonSubsequence = function(text1, text2) { const m = text1.length, n = text2.length; const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0)); for (let i = 1; i <= m; i++) { const c1 = text1[i - 1]; for (let j = 1; j <= n; j++) { const c2 = text2[j - 1]; if (c1 === c2) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); } } } return dp[m][n]; };

[smallppgirl]

思路： DP (二维）

代码

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if text1[i - 1] == text2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        return dp[m][n]

时间复杂度： O(mn) 空间复杂度： O(mn)

[snmyj]

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int len1 = text1.length(), len2 = text2.length();
        vector<vector<int>> lens(len1 + 1, vector<int>(len2 + 1));

        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (text1[i - 1] == text2[j - 1]) {
                    lens[i][j] = lens[i - 1][j - 1] + 1;
                } else {
                    lens[i][j] = max(lens[i - 1][j], lens[i][j - 1]);
                }
            }
        }

        return lens[len1][len2];
    }
};

[joe-the-plumber]

二维DP 打个卡

class Solution:
    def longestCommonSubsequence(self, A: str, B: str) -> int:
        m, n = len(A), len(B)
        ans = 0
        dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if A[i - 1] == B[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                    ans = max(ans, dp[i][j])
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        return ans