Question

步骤1.选择虚拟键。（苹果）
  步骤2.从虚拟键中重复数据删除。（APLE）

步骤3.剩余的字母表连接在一起，因此没有   从最后一个字母表重复。这时，总长度是   26个字符。如果＆＃39; z＆＃39;填补，它从＆＃39; a＆＃39;试。

结果：aplefghijkmnoqrstuvwxyzbcd

修正了第2步，我想在文本字段中使用.getText（）我不知道如何从第3步开始。
我该怎么办？

import java.awt.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.event.WindowEvent;
import java.awt.event.WindowListener;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.Set;

import javax.swing.*;
import javax.swing.border.LineBorder;
import javax.swing.border.TitledBorder;

import com.jgoodies.forms.factories.DefaultComponentFactory;

public class monoalphabetic_Cipher extends JFrame implements ActionListener {

    private JPanel p1, p2, p3;      // default, staging job, job selection
    private JLabel title, l1, l2, l3, l4;   // title, virtual key input, de-duplicated virtual key, encryption, decryption
    private JButton b1, b2, b3;         // Encryption, decryption, deduplication
    private JTextField t1, t2, t3, t4;
    int i,j;

    public monoalphabetic_Cipher() {

        super("monoalphabetical cipher");

        p1 = new JPanel();
        p1.setForeground(new Color(0, 0, 0));
        p1.setBackground(UIManager.getColor("Button.background"));

        title = new JLabel("<html><h1>monoalphabetical cipher</h1><hr></html>");
        title.setFont(new Font("맑은 고딕", Font.PLAIN, 12));
        title.setBounds(204, 20, 175, 57);
        p1.add(title);

        b2 = new JButton("encryption");
        b2.setFont(new Font("맑은 고딕", Font.PLAIN, 12));
        b2.setBackground(UIManager.getColor("Button.background"));
        b2.setBounds(67, 210, 119, 23);
        p1.add(b2);

        b3 = new JButton("decryption");
        b3.setFont(new Font("맑은 고딕", Font.PLAIN, 12));
        b3.setBackground(UIManager.getColor("Button.background"));
        b3.setBounds(67, 252, 119, 23);
        p1.add(b3);

        t3 = new JTextField(20);
        t3.setBounds(260, 212, 297, 21);
        p1.add(t3);

        t4 = new JTextField(20);
        t4.setBounds(260, 254, 297, 21);
        p1.add(t4);

        p1.setLayout(null);

        t2 = new JTextField(10);
        t2.setBounds(395, 123, 128, 21);
        p1.add(t2);

        b1 = new JButton("overlap remove");
        b1.setFont(new Font("맑은 고딕", Font.PLAIN, 12));
        b1.setBounds(264, 122, 119, 23);        
        p1.add(b1);

        l1 = new JLabel("virtual key");
        l1.setFont(new Font("맑은 고딕", Font.PLAIN, 12));
        l1.setBounds(55, 126, 57, 15);
        p1.add(l1);

        t1 = new JTextField(10);
        t1.setBounds(124, 123, 128, 21);
        p1.add(t1);



        p3 = new JPanel();
        p3.setBorder(new TitledBorder(new LineBorder(Color.lightGray), " selection", TitledBorder.LEADING, TitledBorder.TOP, new Font("맑은 고딕", Font.PLAIN, 12), new Color(0, 0, 0)));
        p3.setBounds(34, 176, 184, 122);
        p1.add(p3);

        l2 = new JLabel("\u2460");
        l2.setFont(new Font("맑은 고딕", Font.PLAIN, 16));
        l2.setBounds(230, 211, 28, 23);
        p1.add(l2);

        l3 = new JLabel("\u2461 ");
        l3.setFont(new Font("맑은 고딕", Font.PLAIN, 16));
        l3.setBounds(230, 252, 28, 23);
        p1.add(l3);


        p2 = new JPanel();
        p2.setBorder(new TitledBorder(new LineBorder(Color.lightGray), "ready", TitledBorder.LEADING, TitledBorder.TOP, new Font("맑은 고딕", Font.PLAIN, 12), new Color(0, 0, 0)));
        p2.setBounds(34, 97, 523, 69);
        p1.add(p2);
        getContentPane().add(p1);
        setSize(600,355);
        setResizable(false);                                
        setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE); 
        setVisible(true);


        // duplication remove
        b1.addActionListener(new ActionListener() {
            public void actionPerformed(ActionEvent e) {

                int vk_1 = t1.getText().length();

                Set chk = new LinkedHashSet();

                for(i=0;i<vk_1;i++) {
                    // Convert all lowercase letters to uppercase and lowercase letters
                    chk.add(t1.getText().toLowerCase().charAt(i)); 
                }
                /// Remove brackets and commas in the [result, result] format, which is the default output method of LinkedHashSet()
                String result = Arrays.toString(chk.toArray()).replace("[", "").replace(",", "").replace("]", "");

                / * With the usual replaceAll ("", "") or trim ()
                  * If spaces are not removed, use regular expressions
                  * /tText(result.replaceAll("\\p{Z}", "")); 

                if(t1.getText().equals("")) {
                    t2.setText("no input key");
                }
            }
        });

        // encryption
        b2.addActionListener(new ActionListener() {

            final String curString = t2.getText();

            final char startChar = curString.charAt(curString.length()-1);

            public void actionPerformed(ActionEvent e) {

                StringBuilder sb = new StringBuilder();
                sb.append(curString);

                char c = nextChar(startChar);

                while (sb.length() < 26) {

                    while (hasChar(sb, c)) {
                        c = nextChar(c);
                    }                   
                    sb.append(c);


                    c = nextChar(c);
                }

                System.out.printf("Final String %28s has length of %3d", sb.toString(), sb.length());


            }
            private char nextChar(int idx) {
                 ++idx;
                    if (idx > 'z') {
                        idx = 'a';
                    }

                    return (char)idx;
            }
            private boolean hasChar(StringBuilder sb, char c)
            {
                for (int i = 0; i < sb.length(); ++i) {
                    if (sb.charAt(i) == c) {
                        return true;
                    }
                }

                return false;
            }
        });

        // Decrypt
        b3.addActionListener(new ActionListener() {
            public void actionPerformed(ActionEvent e) {
                if(t2.getText().equals("")) {
                    t4.setText("no input key");
                }
            }
        });
    }   

    public static void main(String[] args) {    
        new monoalphabetic_Cipher();
    }
}

Answer 1

假设：输入字符串经过重复数据删除，它具有适当的长度，要添加的字符为“a”到“z”（包括端点），不应重复任何字符，最终字符串应为26个字符长，以下提供了一个快速的解决方案。需要进行额外的错误检查和更好的变量命名。

public static void main(String[] args)
{
    // the example string
    final String curString = "aple";

    // get the starting character, which is the end of the string
    // in the example, will be 'e'
    final char startChar = curString.charAt(curString.length() - 1);

    // construct our compilation; this could also be List<Character>
    //  which would make somethings easier, but would have to assemble
    //  at the end
    StringBuilder sb = new StringBuilder();
    sb.append(curString);

    // we want the character after the one on the end of the string to start
    char c = nextChar(startChar);        

    // we want 26 total entries
    while (sb.length() < 26) {

        // we cannot append a character that already exists, so spin
        // until we find one not in the compilation
        while (hasChar(sb, c)) {
            c = nextChar(c);
        }

        // we now have the one we want, so append it
        sb.append(c);

        // we know we need at least the next character after the one we
        //  just appended, so get it and loop; loop will terminate if
        //  we've added enough
        c = nextChar(c);
    }


    // display the result
    System.out.printf("Final String %28s has length of %3d", sb.toString(), 
            sb.length());
}


private static boolean hasChar(StringBuilder sb, char c)
{
    for (int i = 0; i < sb.length(); ++i) {
        if (sb.charAt(i) == c) {
            return true;
        }
    }

    return false;
}

private static char nextChar(int idx)
{
    ++idx;
    if (idx > 'z') {
        idx = 'a';
    }

    return (char)idx;
}

输出：

Final String aplefghijkmnoqrstuvwxyzbcd的长度为26

java swing中的单字母加密/解密

1 个答案: