1143. 最长公共子序列

解法一：动态规划

dp[i][j] 表示 text1[0:i] 和 text2[0:j] 的最长公共子序列长度。状态转移方程如下：

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        char[] s1 = text1.toCharArray();
        char[] s2 = text2.toCharArray();
        final int len1 = s1.length;
        final int len2 = s2.length;
        // dp[i][j]表示text1[0:i]和text2[0:j]的最长公共子序列长度
        int[][] dp = new int[len1 + 1][len2 + 1];
        for (int i = 1; i <= len1; ++i) {
            for (int j = 1; j <= len2; ++j) {
                if (s1[i - 1] == s2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[len1][len2];
    }
}