2022.05.26(LC

①确定状态：dp[i][j]表示长度[0, i - 1]的字符串text1与长度[0, j - 1]的字符串text2的最长公共子序列；

②转移方程：dp[i][j]=dp[i−1][j−1]+1, 当 text1[i - 1] == text2[j - 1];

                     dp[i][j]=Math.max(dp[i−1][j],dp[i][j−1]), 当 text1[i - 1] != text2[j - 1];

③初始条件和边界情况：dp[i][0] = 0;dp[0][j] = 0;

④计算顺序：因为 dp[i][j] 由 dp[i−1][j−1]，dp[i - 1][j] 和 dp[i][j - 1]推出，所以从小到大遍历。

长公共子序列题解https://leetcode.cn/problems/longest-common-subsequence/solution/fu-xue-ming-zhu-er-wei-dong-tai-gui-hua-r5ez6/

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length(), n = text2.length();
        //dp[i][j]表示长度[0, i - 1]的字符串text1与长度[0, j - 1]的字符串text2的最长公共子序列
        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)) {
                    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[m][n];
    }
}