[LeetCode] 1143. Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

• For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

Example 1:

Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

• 1 <= text1.length, text2.length <= 1000
• text1 and text2 consist of only lowercase English characters.

最长公共子序列。

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。如果不存在 公共子序列 ，返回 0 。

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。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/longest-common-subsequence
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这个题可以跟718题一起刷，基本是一样的意思，思路都是动态规划的背包问题。

首先创建DP数组，dp[i][j] 的含义是以 text1 在位置 i 和 text2 在位置 j 为结尾的最长的公共子序列的长度。初始化的时候 dp[0][0] = 0

如果当前的字母相同，text1.charAt(i - 1) == text2.charAt(j - 1)，那么dp[i][j] = dp[i - 1][j - 1] + 1

否则，dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])，

此时最长公共子序列为 text1 的前 i - 1 个字符和 text2 的前 j 个字符最长公共子序列，或者 text1 的前 i 个字符和 text2 的前 j - 1 个字符最长公共子序列，取它们之间较大的一个

时间O(mn)

空间O(mn)

Java实现

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

 

JavaScript实现

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function (text1, text2) {
    let m = text1.length;
    let n = text2.length;
    let dp = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            dp[i][j] =
                text1[i - 1] == text2[j - 1]
                    ? dp[i - 1][j - 1] + 1
                    : Math.max(dp[i - 1][j], dp[i][j - 1]);
        }
    }
    return dp[m][n];
};

 

相关题目

718. Maximum Length of Repeated Subarray

1143. Longest Common Subsequence

LeetCode 题目总结