【LeetCode】 1143. Longest Common Subsequence

Description

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

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. (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.

Constraints:

• 1 <= text1.length <= 1000
• 1 <= text2.length <= 1000
• The input strings consist of lowercase English characters only.

給兩個字串 text1 和 text2, 回傳它們最長共同子字串的長度。

一個字串的子字串是從原始字串生成的新字串，其中刪除了一些字元(可以是空字元)而不改變其餘字元的相對順序。(例如："ace"是"abcde"的子字串，而"aec"則不是)。兩個字串的共同子字串是指該子字串同時為兩者的子字串。

如果沒有任何共同子字串，回傳0。

限制：

• 1 <= text1.length <= 1000
• 1 <= text2.length <= 1000
• 輸入字串只會包含小寫英文字母。

Example:

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.

Solution

• 經典的DP題目。
• 首先我們設兩個變數m和n分別為兩個字串的長度。
• 建一個DP表，且大小為m + 1 x n + 1。
  • 表格的意義為：text1和text2只取前i個和前j個字的時候的最長共同子字串。
  • DP表大小要多一格是因為要考慮到空字串。
• 空字串一定沒有共同子字串(或者說共同子字串只有空字串)，因此表格中i = 0或是j = 0都填入0。
• 當text[i] == text[j]時，此時填入dp[i - 1][j - 1] + 1。
  • 代表目前最長的就是去掉最後一個字的最長子字串再 +1。
• 若是不等於，填入dp[i - 1][j]和dp[i][j - 1]較大者。
  • 代表不看text1或是text2的最後一個字，取較長者。
• 整理成遞迴式：
  
  $$\begin{gathered} dp[i][j]= \\ \{\begin{array}{c}0,i=0ORj=0\\ dp[i-1][j-1],s[i]=s[j]\\ max(dp[i-1][j],dp[i][j-1]),other\end{array} \end{gathered}$$
• 最後取dp[m][n]就是答案了。
  • 前m個和前n個就是原來的兩個字串。

Code

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.length();
        int n = text2.length();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        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];
    }
};

tags: LeetCode C++