1143. Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence.

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. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A  common subsequence  of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

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 <= 1000
  • 1 <= text2.length <= 1000
  • The input strings consist of lowercase English characters only.

这道题让求最长相同的子序列,注意是子序列,不是子串,所以字符并不需要相连,但是字符顺序还是需要保持的。LeetCode 之前也有题目需要借助求 LCS 来解题,比如 Delete Operation for Two Strings,当时博主还疑惑怎么 LeetCode 中没有专门求 LCS 的题呢,这不,终于补上了。解题思路和上面那道是一模一样,若搞懂了那道题,这道也就是没什么难度了,这里是用动态规划 Dynamic Programing 来做,使用一个二维数组 dp,其中 dp[i][j] 表示 text1 的前i个字符和 text2 的前j个字符的最长相同的子序列的字符个数,这里大小初始化为 (m+1)x(n+1),这里的m和n分别是 text1 和 text2 的长度。接下来就要找状态转移方程了,如何来更新 dp[i][j],若二者对应位置的字符相同,表示当前的 LCS 又增加了一位,所以可以用 dp[i-1][j-1] + 1 来更新 dp[i][j]。否则若对应位置的字符不相同,由于是子序列,还可以错位比较,可以分别从 text1 或者 text2 去掉一个当前字符,那么其 dp 值就是 dp[i-1][j] 和 dp[i][j-1],取二者中的较大值来更新 dp[i][j] 即可,最终的结果保存在了 dp[m][n] 中,参见代码如下:

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size(), n = text2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                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];
    }
};

讨论:这道题跟之前那道 Longest Palindromic Subsequence 的解题思路几乎是一模一样,你品,你细品,一定要通过现象看本质,才能举一反三,触类旁通。

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1143

类似题目:

Longest Palindromic Subsequence

Delete Operation for Two Strings

Shortest Common Supersequence

参考资料:

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

https://leetcode.com/problems/longest-common-subsequence/discuss/348884/C%2B%2B-with-picture-O(nm)

https://leetcode.com/problems/longest-common-subsequence/discuss/351689/JavaPython-3-Two-DP-codes-of-O(mn)-and-O(min(m-n))-spaces-w-picture-and-analysis

LeetCode All in One 题目讲解汇总(持续更新中…)


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