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

Question

difficulty: mid
adj diff: 2

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.

DP，不难。

Code

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public int longestCommonSubsequence(String text1, String text2) {
    int m = text1.length();
    int n = text2.length();
    int[][] dp = new int[m][n];

    char[] s1 = text1.toCharArray();
    char[] s2 = text2.toCharArray();
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            // dp[i][j] represent the LCS up till s1[i] and s2[j] inclusive
            // if char is same, dp[i][j] = 1 + dp[i-1][j-1]
            // else, dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1])
            if (i == 0 && j == 0) {
                dp[i][j] = s1[i] == s2[j] ? 1 : 0;
            } else if (i == 0) {
                dp[i][j] = s1[i] == s2[j] ? 1 : dp[i][j-1];
            } else if (j == 0) {
                dp[i][j] = s1[i] == s2[j] ? 1 : dp[i-1][j];
            } else {
                dp[i][j] = s1[i] == s2[j] ? 1 + dp[i-1][j-1] : Math.max(dp[i-1][j], dp[i][j-1]);
            }
        }
    }
    return dp[m - 1][n - 1];
}