1143 longest common subsequence

Description

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.

Examples

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.

Ideas

dp, Construct a 2 D matrix ,dp[i][j] Express text1.substring(0, i + 1) and text2.substring(0, j + 1) There are several identical elements , This update dp[i][j] The state transition formula is

• If text1.charAt(i) == text1.charAt(j)——dp[i][j] = dp[i-1][j-1] + 1
• The second step , perform dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1], dp[i][j])

Code

class Solution {
    
    public int longestCommonSubsequence(String text1, String text2) {
    
        int[][] dp = new int[text1.length() + 1][text2.length() + 1];
        
        for (int i = 0; i < text1.length(); i++) {
    
            for (int j = 0; j < text2.length(); j++) {
    
                int add = text1.charAt(i) == text2.charAt(j)? 1: 0;
                dp[i + 1][j + 1] = dp[i][j] + add;
                dp[i + 1][j + 1] = Math.max(dp[i][j + 1], dp[i + 1][j + 1]);
                dp[i + 1][j + 1] = Math.max(dp[i + 1][j], dp[i + 1][j + 1]);
            }
        }
        
        return dp[text1.length()][text2.length()];
    }
}