第二题：给你一串由小写字母构成的字符串，若字符串中两个字符相同，你可以将其标记并得分，得分为两个字符在字母表中的位置（从1开始）之和；若两个字符在字母表中相邻，也可以将其标记，例如ab相邻，yx相邻。每个字符只能被标记一次，返回所能取得的最高分。

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        int n = s.length();
        if (n < 2) {
            System.out.println(0);
            return;
        }
        // dp[i]表示以i为结尾的最大贡献
        int[] dp = new int[n];
        dp[0] = 0;
        dp[1] = cal(s.charAt(0), s. charAt(1));
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i-2] + cal(s.charAt(i-1), s.charAt(i)), dp[i-1]);
        }
        System.out.println(dp[n-1]);
    }

    public static int cal(char a, char b) {
        int x = a - 'a' + 1;
        int y = b - 'a' + 1;
        if (Math.abs(x-y) <= 1) {
            return x + y;
        }
        return 0;
    }
}

 第三题：给你一个数字n，让你将数字1-n填入完全二叉树，要求除根节点外，每个节点值与父节点值的乘积是偶数。

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        System.out.print(2);
        for (int i = 4; i <= n; i += 2) {
            System.out.print(" " + i);
        }
        for (int i = 1; i <= n; i += 2) {
            System.out.print(" " + i);
        }
    }
}

第四题：二维dp 求 左上角 到 右下角的最短路径 平原到平原、沼泽到沼泽的cost为1；平原到沼泽和沼泽到平原的cost为2

// 二维dp 向下 向右走即可

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        int[][] board = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                board[i][j] = sc.nextInt();
            }
        }

        int[][] dp = new int[n][m];
        dp[0][0] = 0;
        for (int i = 1; i < n; i++) {
            if (board[i][0] == board[i-1][0]) {
                dp[i][0] = dp[i-1][0] + 1;
            } else {
                dp[i][0] = dp[i-1][0] + 2;
            }
        }

        for (int j = 1; j < m; j++) {
            if (board[0][j] == board[0][j-1]) {
                dp[0][j] = dp[0][j-1] + 1;
            } else {
                dp[0][j] = dp[0][j-1] + 2;
            }
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                dp[i][j] = Math.min(dp[i-1][j] + (board[i-1][j] == board[i][j]? 1: 2),
                        dp[i][j-1] + (board[i][j-1] == board[i][j]? 1: 2));
            }
        }
        System.out.println(dp[n-1][m-1]);
    }
}