1143. Longest Common Subsequence 

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class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length(), n = text2.length();

        // 定义 dp[i][j] 为 text1 前 i 个字符组成的字符串与 text2 前 j 个字符组成的字符串的最长公共子序列的长度
        int[][] dp = new int[m + 1][n + 1];

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1.charAt(i - 1) == text2.charAt(j - 1)) { // 注意此处的索引
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }

        return dp[m][n];
    }
}

References

1143. Longest Common Subsequence
剑指 Offer II 095. 最长公共子序列