.import java.awt.*;import java.awt.event.*;import javax.swing.*;import javax.swing.table.DefaultTableModel;import java.sql.*;import java.util.Vector;publicclass LL1 extends JFrame implements ActionListener {/** * */privatestaticfinallongserialVersionUID = 1L; JTextField tf1; JTextField tf2; JLabel l; JButton b0; JPanel p1, p2, p3; JTextArea t1, t2, t3; JButton b1, b2, b3; JLabel l0, l1, l2, l3, l4; JTable table; Statement sta; Connection conn; ResultSet rs; DefaultTableModel dtm; String Vn[] = null; Vector P = null;intfirstComplete[] = null;// 存储已判断过first的数据charfirst[][] = null;// 存储最后first结果intfollowComplete[] = null;// 存储已判断过follow的数据charfollow[][] = null;// 存储最后follow结果charselect[][] = null;// 存储最后select结果intLL = 0;// 标记是否为LL〔1〕 String vt_tou[] = null;// 储存Vt Object shuju[][] = null;// 存储表达式数据charyn_null[] = null;// 存储能否推出空 LL1() { setLocation(100, 0); setSize(700, 780);tf1 = new JTextField(13);tf2 = new JTextField(13);l = new JLabel(">>");l0 = new JLabel("输入字符串：");l1 = new JLabel("输入的文法为： ");l2 = new JLabel(" ");l3 = new JLabel("分析的结果： ");l4 = new JLabel("预测分析表： ");// p1=new JPanel();p2 = new JPanel();p3 = new JPanel();t1 = new JTextArea(24, 20);t2 = new JTextArea(1, 30);t3 = new JTextArea(24, 40);b0 = new JButton("确定(S为开场)");b1 = new JButton(" 判断文法 ");b2 = new JButton("输入");b3 = new JButton("清空");table = new JTable(); JScrollPane jp1 = new JScrollPane(t1); JScrollPane jp2 = new JScrollPane(t2); JScrollPane jp3 = new JScrollPane(t3);p2.add(tf1);p2.add(l);p2.add(tf2);p2.add(b0);p2.add(b1);p2.add(l0);p2.add(l2);p2.add(jp2);p2.add(b2);p2.add(b3);p2.add(l1);p2.add(l3);p2.add(jp1);p2.add(jp3);p3.add(l4);p3.add(new JScrollPane(table)); add(p2, "Center"); add(p3, "South");b0.addActionListener(this);b1.addActionListener(this);b2.addActionListener(this);b3.addActionListener(this); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);table.setPreferredScrollableViewportSize(new Dimension(660, 200)); setVisible(true); }////////////////////界面设计 publicvoid actionPerformed(ActionEvent e) {if (e.getSource() == b0) { String a = tf1.getText(); String b = tf2.getText();t1.append(a + '→' + b + '\n'); }if (e.getSource() == b1) {t3.setText(");intVnnum = 0, k;Vn = new String[100];P = new Vector(); String s[] = t1.getText().split("\n");for (inti = 0; i < s.length; i++) {if (s.length < 2) {t3.setText("文法输入有误，请重新输入");// 判断长度是否符合return; }if (s[i].charAt(0) <= 'Z' && s[i].charAt(0) >= 'A' && s[i].charAt(1) == '→') {for (k = 0; k < Vnnum; k++) {if (Vn[k].equals(s[i].substring(0, 1))) {break; } }if (Vnnum == 0 || k >= Vnnum) {Vn[Vnnum] = s[i].substring(0, 1);// 存入Vn数据Vnnum++; }P.add(s[i]); } else {t3.setText("文法输入有误，请重新输入");return; } }yn_null = newchar[100];first = newchar[Vnnum][100];intflag = 0; String firstVn[] = null;firstComplete = newint[Vnnum];for (inti = 0; Vn[i] != null; i++) // 依次求 FIRST** {flag = 0;firstVn = new String[20];if ((flag = add_First(first[i], Vn[i], firstVn, flag)) == -1)return;firstComplete[i] = 1; }t3.append("first集：" + "\n"); // 显示FIRST**for (inti = 0; Vn[i] != null; i++) {t3.append("first(" + Vn[i] + ")={ ");for (intj = 0; first[i][j] != '\0'; j++) {t3.append(first[i][j] + " , "); }t3.append("}" + "\n"); }follow = newchar[Vnnum][100]; String followVn[] = null;followComplete = newint[Vnnum];for (inti = 0; Vn[i] != null; i++) // 求FOLLOW** {flag = 0;followVn = new String[20];if ((flag = tianjiaFollow(follow[i], Vn[i], followVn, flag)) == -1)return;followComplete[i] = 1; }t3.append("follow集：" + "\n"); // 显示FOLLOW**for (inti = 0; Vn[i] != null; i++) {t3.append("follow(" + Vn[i] + ")={ ");for (intj = 0; follow[i][j] != '\0'; j++) {t3.append(follow[i][j] + " , "); }t3.append("}" + "\n"); }select = newchar[P.size()][100];for (inti = 0; i < P.size(); i++) // 求SELECT** {flag = 0; tianjiaSelect(select[i], (String) P.elementAt(i), flag); }t3.append("select集：" + "\n"); // 显示SELECT**for (inti = 0; i < P.size(); i++) {t3.append("select(" + (String) P.elementAt(i) + ")={ ");for (intj = 0; select[i][j] != '\0'; j++) {t3.append(select[i][j] + " , "); }t3.append("}" + "\n"); }for (inti = 0; Vn[i] != null; i++)// 判断select交集是否为空 {intbiaozhi = 0;charsave[] = newchar[100];for (intj = 0; j < P.size(); j++) { String t = (String) P.elementAt(j);if (t.substring(0, 1).equals(Vn[i])) {for (k = 0; select[j][k] != '\0'; k++) {if (puanduanChar(save, select[j][k])) {save[biaozhi] = select[j][k];biaozhi++; } else// 当有交集时，不为LL〔1〕文法 {t3.append("不是LL〔1〕文法！！" + "\n");return; } } } } }charVt[] = newchar[100];intbiaozhi = 0;for (inti = 0; i < P.size(); i++) { String t = (String) P.elementAt(i);for (intj = 2; j < t.length(); j++)// 提取表达式右侧的终结符存入Vt {if (t.charAt(j) > 'Z' || t.charAt(j) < 'A') {if (puanduanChar(Vt, t.charAt(j))) {Vt[biaozhi] = t.charAt(j);biaozhi++; } } } }if (puanduanChar(Vt, '*'))// 假设可推出空集，那么将*参加Vt。
{Vt[biaozhi] = '*';biaozhi++; }vt_tou = new String[biaozhi + 1];// 根据select和表达式生成预测分析表shuju = new String[Vnnum][biaozhi + 1]; String f = ";vt_tou[0] = f;for (inti = 0; i < biaozhi; i++) {vt_tou[i + 1] = String.valueOf(Vt[i]); }for (inti = 0; i < Vnnum; i++) {shuju[i][0] = Vn[i]; }for (inti = 0; i < P.size(); i++) { String t = (String) P.elementAt(i);intj;for (j = 0; j < Vn.length; j++) {if (Vn[j].equals(t.substring(0, 1))) {break; } }for (k = 0; select[i][k] != '\0'; k++) {inty = puanduanXulie(Vt, select[i][k]);shuju[j][y + 1] = t.substring(1); } }dtm = new DefaultTableModel(shuju, vt_tou);// 显示预测分析表table.setModel(dtm);LL = 1; }if (e.getSource() == b3)// 清空列表 {tf1.setText(");tf2.setText(");t1.setText(");t2.setText(");t3.setText(");Vn = null;P = null;firstComplete = null;first = null;followComplete = null;follow = null;select = null;dtm = new DefaultTableModel();table.setModel(dtm); }if (e.getSource() == b2)// 输入字符串并预测分析 {t3.setText(");if (LL == 1) { String s = t2.getText();for (inti = 0; i < s.length(); i++) {if (s.charAt(i) == '\0') {t3.setText("字符串中请不要参加空格" + "\n");return; } }charzifu[] = newchar[100];// 剩余输入串charfenxi[] = newchar[100];// 分析栈zifu[0] = '*';fenxi[1] = 'S';fenxi[0] = '*';intfzifu = 1;intffenxi = 2;for (inti = s.length() - 1; i >= 0; i--) {zifu[fzifu] = s.charAt(i);fzifu++; }intbuzhou = 2;charn[] = newchar[65];// 存储要显示的容t3.append("步骤 分析栈 剩余输入串 所用产生式或匹配" + "\n");n[0] = '1';n[15] = '*';n[14] = 'S';intu = 29;for (inti = fzifu - 1; i >= 0; i--) {n[u] = zifu[i];u++; }while (!(fenxi[ffenxi - 1] == '*' && zifu[fzifu - 1] == '*'))// 剩余输入串不为*那么分析 {inti, j;chart = zifu[fzifu - 1];chark = fenxi[ffenxi - 1];if (t == k)// 产生式匹配 {n[49] = k;n[50] = '匹';n[51] = '配';t3.append(String.copyValueOf(n) + "\n");n = newchar[65];fzifu--;ffenxi--;if (buzhou < 10)n[0] = (char) ('0' + buzhou);// 显示步骤数else {n[0] = (char) ('0' + buzhou / 10);n[1] = (char) ('0' + buzhou % 10); }u = 14;for (inty = ffenxi - 1; y >= 0; y--)// 处理分析栈，出栈 {n[u] = fenxi[y];u++; }u = 29;for (inty = fzifu - 1; y >= 0; y--)// 处理剩余字符串，消除一个字符 {n[u] = zifu[y];u++; }buzhou++;continue; }for (i = 0; Vn[i] != null; i++)// 搜寻所用产生式的左部 {if (Vn[i].equals(String.valueOf(k)))break; }for (j = 0; j < vt_tou.length; j++)// 判断是否匹配 {if (vt_tou[j].equals(String.valueOf(t)))break; }if (j >= vt_tou.length)// 全部产生式都不能符合那么报错 {t3.append(String.copyValueOf(n));t3.append("\n" + "该串不是该文法的句型" + "\n");return; } Object result1 = shuju[i][j];if (result1 == null) {t3.append(String.copyValueOf(n));t3.append("\n" + "该串不是该文法的句型" + "\n");return; } else// 找到所用产生式 {n[49] = Vn[i].charAt(0);u = 50; String result = (String) result1;for (inty = 0; y < result.length(); y++) {n[u] = result.charAt(y);u++; }t3.append(String.copyValueOf(n) + "\n");n = newchar[65];ffenxi--;for (i = result.length() - 1; i > 0; i--)// 将分析栈非终结符换为右边表达式 {if (result.charAt(i) != '*') {fenxi[ffenxi] = result.charAt(i);ffenxi++; } } }if (buzhou < 10)// 显示"步骤〞n[0] = (char) ('0' + buzhou);else {n[0] = (char) ('0' + buzhou / 10);n[1] = (char) ('0' + buzhou % 10); }u = 14;for (inty = ffenxi - 1; y >= 0; y--) {n[u] = fenxi[y];u++; }u = 29;for (inty = fzifu - 1; y >= 0; y--) {n[u] = zifu[y];u++; }buzhou++; }n = newchar[65];n[0] = '1';n[14] = '*';n[29] = '*';n[49] = '分';n[50] = '析';n[51] = '成';n[52] = '功';t3.append(String.copyValueOf(n));t3.append("\n" + "该串是此文法的句型" + "\n");return; } else {t3.setText("请先依次输入LL〔1〕文法，并点击 文法判断 按钮" + "\n");return; } } }privateint add_First(chara[], String b, String firstVn[], intflag) // 计算FIRST**〔递归〕 {if (puanduanString(firstVn, b.charAt(0))) { addString(firstVn, b); } else {returnflag; }for (inti = 0; i < P.size(); i++) { String t = (String) P.elementAt(i);for (intk = 2; k < t.length(); k++) {if (t.substring(0, 1).equals(b)) {if (t.charAt(k) > 'Z' || t.charAt(k) < 'A')// 表达式右端第i个不是非终结符 {if (flag == 0 || puanduanChar(a, t.charAt(k))) {if (t.charAt(k) == '*')// *时直接参加first {if (k + 1 == t.length()) {a[flag] = t.charAt(k);flag++; }intflag1 = 0;for (intj = 0; yn_null[j] != '\0'; j++)// 所求非终结符进入yn_null** {if (yn_null[j] == b.charAt(0))// 判断能否推出空 {flag1 = 1;break; } }if (flag1 == 0) {intj;for (j = 0; yn_null[j] != '\0'; j++) { }yn_null[j] = b.charAt(0); }continue; } else// 终结符直接参加first** {a[flag] = t.charAt(k);flag++;break; } }break; } else// 非终结符 {if (!puanduanString(Vn, t.charAt(k))) {intp = firstComplete(t.charAt(k));// 当该非终结符的first已经求出if (p != -1) {flag = addElementFirst(a, p, flag);// 直接参加所求first } elseif ((flag = add_First(a, String.valueOf(t.charAt(k)), firstVn, flag)) == -1)return -1;intflag1 = 0;for (intj = 0; yn_null[j] != '\0'; j++)// 当非终结符的first有空 {if (yn_null[j] == t.charAt(k)) {flag1 = 1;break; } }if (flag1 == 1)// 当非终结符的first能推出空 {if (k + 1 == t.length() && puanduanChar(a, '*'))// 之后无符号，且所求first无* {a[flag] = '*';// first中参加*flag++; }continue;// 判断下一个字符 } else {break; } } else// 错误 {t3.setText("文法输入有误" + "\n");return -1; } } } } }returnflag; }privateint tianjiaFollow(chara[], String b, String followVn[], intflag) // 计算FOLLOW**〔递归〕 {if (puanduanString(followVn, b.charAt(0))) { addString(followVn, b); } else {returnflag; }if (b.equals("S"))// 当为S时*存入follow {a[flag] = '*';flag++; }for (inti = 0; i < P.size(); i++) { String t = (String) P.elementAt(i);for (intj = 2; j < t.length(); j++) {if (t.charAt(j) == b.charAt(0) && j + 1 < t.length()) {if (t.charAt(j + 1) != '\0') {if ((t.charAt(j + 1) > 'Z' || t.charAt(j + 1) < 'A'))// 所求为终结符 {if (flag == 0 || puanduanChar(a, t.charAt(2)))// 自身 {a[flag] = t.charAt(j + 1);flag++; } }else// 所求为非终结符 {intk;for (k = 0; Vn[k] != null; k++) {if (Vn[k] .equals(String.valueOf(t.charAt(j + 1)))) {break;// 找寻下一个非终结符的Vn位置 } }flag = addElementFirst(a, k, flag);// 把下一个非终结符first参加所求follow集for (k = j + 1; k < t.length(); k++) {if ((t.charAt(j + 1) > 'Z' || t.charAt(j + 1) < 'A'))break;else {if (panduan_kong(t.charAt(k))) { } else {break; } } }if (k >= t.length())// 下一个非终结符可推出空，把表达式左边非终结符的follow集参加所求follow集 {intp = followComplete(t.charAt(0));if (p != -1) {flag = addElementFollow(a, p, flag); } elseif ((flag = tianjiaFollow(a, String.valueOf(t.charAt(0)), followVn,flag)) == -1)return -1; } } } else// 错误 {t3.setText("文法输入有误,请重新输入" + "\n");return -1; } }if (t.charAt(j) == b.charAt(0) && j + 1 == t.length())// 下一个字符为空，把表达式左边非终结符的follow集参加所求follow集 {intp = followComplete(t.charAt(0));if (p != -1) {flag = addElementFollow(a, p, flag); } elseif ((flag = tianjiaFollow(a, String.valueOf(t.charAt(0)), followVn, flag)) == -1)return -1; } } }returnflag; }privatevoid tianjiaSelect(chara[], String b, intflag) // 计算SELECT** {inti = 2;intbiaozhi = 0;while (i < b.length()) {if ((b.charAt(i) > 'Z' || b.charAt(i) < 'A') && b.charAt(i) != '*')// 是终结符 {a[flag] = b.charAt(i);// 将这个字符参加到Select**，完毕Select**的计算break; } elseif (b.charAt(i) == '*')// 是空 {intj;for (j = 0; Vn[i] != null; j++)// 将表达式左侧的非终结符的follow参加select {if (Vn[j].equals(b.substring(0, 1))) {break; } }for (intk = 0; follow[j][k] != '\0'; k++) {if (puanduanChar(a, follow[j][k])) {a[flag] = follow[j][k];flag++; } }break; } elseif (b.charAt(i) >= 'A' && b.charAt(i) <= 'Z' && i + 1 < b.length())// 是非终结符且有下一个字符 {intj;for (j = 0; Vn[i] != null; j++) {if (Vn[j].equals(String.valueOf(b.charAt(i)))) {break; } }for (intk = 0; first[j][k] != '\0'; k++) {if (puanduanChar(a, first[j][k]))// 把表达式右侧所有非终结符的first集参加。
{if (first[j][k] == '*')// first中存在空 {biaozhi = 1; } else {a[flag] = first[j][k];flag++; } } }if (biaozhi == 1)// 把右侧所有非终结符的first中的*去除 {i++;biaozhi = 0;continue; } else {biaozhi = 0;break; } } elseif (b.charAt(i) >= 'A' && b.charAt(i) <= 'Z' && i + 1 >= b.length())// 是非终结符且没有下一个字符 {intj;for (j = 0; Vn[i] != null; j++) {if (Vn[j].equals(String.valueOf(b.charAt(i)))) {break; } }for (intk = 0; first[j][k] != '\0'; k++) {if (puanduanChar(a, first[j][k])) {if (first[j][k] == '*') {biaozhi = 1;// 表达式右侧能推出空，标记 } else {a[flag] = first[j][k];// 不能推出空，直接将first集参加select集flag++; } } }if (biaozhi == 1)// 表达式右侧能推出空 {for (j = 0; Vn[i] != null; j++) {if (Vn[j].equals(b.substring(0, 1))) {break; } }for (intk = 0; follow[j][k] != '\0'; k++) {if (puanduanChar(a, follow[j][k])) {a[flag] = follow[j][k];// 将将表达式左侧的非终结符的follow参加selectflag++; } }break; } else {biaozhi = 0;break; } } } }// 返回b在Vt[]的位置privateint puanduanXulie(charVt[], charb) {inti;for (i = 0; Vt[i] != '\0'; i++) {if (Vt[i] == b)break; }returni; }// 判断b是否在a中，在返回false，不在返回trueprivateboolean puanduanChar(chara[], charb) {for (inti = 0; a[i] != '\0'; i++) {if (a[i] == b)returnfalse; }returntrue; }// 判断b是否在a中，在返回false，不在返回trueprivateboolean puanduanString(String a[], charb) {for (inti = 0; a[i] != null; i++) {if (a[i].equals(String.valueOf(b)))returnfalse; }returntrue; }// 把b参加字符串组firstVn[]privatevoid addString(String firstVn[], String b) {inti;for (i = 0; firstVn[i] != null; i++) { }firstVn[i] = b; }// 判断b是否已完成first判断privateint firstComplete(charb) {inti;for (i = 0; Vn[i] != null; i++) {if (Vn[i].equals(String.valueOf(b))) {if (firstComplete[i] == 1)returni;elsereturn -1; } }return -1; }// 判断b是否已完成follow判断privateint followComplete(charb) {for (inti = 0; Vn[i] != null; i++) {if (Vn[i].equals(String.valueOf(b))) {if (followComplete[i] == 1)returni;elsereturn -1; } }return -1; }// 把相应终结符添加到first**中privateint addElementFirst(chara[], intp, intflag) {for (inti = 0; first[p][i] != '\0'; i++) {if (puanduanChar(a, first[p][i]) && first[p][i] != '*') {a[flag] = first[p][i];flag++; } }returnflag; }// 把相应终结符添加到follow**中privateint addElementFollow(chara[], intp, intflag) {for (inti = 0; follow[p][i] != '\0'; i++) {if (puanduanChar(a, follow[p][i])) {a[flag] = follow[p][i];flag++; } }returnflag; }// 判断a能是否包含空privateboolean panduan_kong(chara) {inti;for (i = 0; Vn[i] != null; i++) {if (Vn[i].equals(String.valueOf(a))) {break; } }for (intj = 0; first[i][j] != '\0'; j++) {if (first[i][j] == '*')returntrue; }returnfalse; }publicstaticvoid main(String[] args) {new LL1(); }}教育之通病是教用脑的人不用手，不教用手的人用脑，所以一无所能。
教育革命的对策是手脑联盟，结果是手与脑的力量都可以大到不可思议教育之通病是教用脑的人不用手，不教用手的人用脑，所以一无所能教育革命的对策是手脑联盟，结果是手与脑的力量都可以大到不可思议教育之通病是教用脑的人不用手，不教用手的人用脑，所以一无所能教育革命的对策是手脑联盟，结果是手与脑的力量都可以大到不可思议。