Question

这是一个面试问题。它期望创建用户提供的字符串池。然后他应该能够在输入的字符串上执行Undo（删除最后输入的字符串）和Redo（获取最后删除的字符串）操作。这应该仅使用核心Java数据结构来完成。怎么做？

Answer 1

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Stack;

public class RedoUndo {

    static Map<Integer, String> m = new HashMap<>();
    static Stack<String> stack = new Stack<>();
    static int size = 0;

    public static void main(String arg[]) {

        Scanner in = new Scanner(System.in);
        System.out.println("Numer of string you need");
        int n = in.nextInt();
        System.out.println("Enter Strings");
        for (int i = 0; i < n; i++) {
            size++;
            m.put(i, in.next());
        }
        size = size - 1;
        System.out.println("orignal : " + m); // 1
        undo();
        undo();
        redo();
        redo();
    }

    private static void undo() {
        // TODO Auto-generated method stub
        removeMy();
        System.out.println("undo : " + m);// 2
    }

    private static void redo() {
        // TODO Auto-generated method stub
        addMy();
        System.out.println("redo : " + m);// 3
    }

    public static void addMy() {
        String s = stack.pop();
        // TODO Auto-generated method stub
        size++;
        m.put(size, s);

    }

    public static void removeMy() {

        // TODO Auto-generated method stub

        String str = m.get(size);
        m.remove(size);
        stack.add(str);
        size--;

    }

}

Answer 2

这个问题太广泛，可以有很多不同的方法。我想，这很容易。

import java.util.Scanner;
import java.util.Stack;
import java.util.Arrays;

public class RedoUndo {
    static Stack<String> stack1 = new Stack<>(); // your current data
    static Stack<String> stack2 = new Stack<>(); // contains strings that you "undo"

    public static void main(String arg[])
    {
        Scanner in = new Scanner(System.in);
        while(true)
        {
            System.out.println("press 1:enter new  2:undo  3:redo");
            switch(in.nextInt())
            {
                case 1:
                System.out.println("enter it:");
                stack1.push(in.next());
                break;

                case 2:
                stack2.push(stack1.pop());
                break;

                case 3:
                stack1.push(stack2.pop());
                break;
            }

            System.out.println("current data:" + Arrays.toString(stack1.toArray()));
            System.out.println("data for redo:" + Arrays.toString(stack2.toArray()));
        }
    }
}

如何对字符串执行撤消和重做操作？

2 个答案: