一、实验目的和要求：

1． 掌握语法分析的基本思想，并用高级语言编写逆波兰式生成程序

2． 要求利用逆波兰式生成算法编写程序，将从键盘上输入的算术表达式

(中缀表达式)转化成逆波兰式

二、实验平台:

Java语言

三、主要实验内容及结果:

实验内容：

逆波兰表达式是算术表达式的后缀表示形式，逆波兰表达式生成算法的关 键在于比较当前运算符与栈顶运算符的优先关系，若当前运算符的优先级高于 栈顶运算符，则当前运算符入栈，若当前运算符的优先级低于栈顶运算符，则 栈顶运算符退栈。

程序代码：

import java.util.ArrayList;

public class InversePolish {// 存放运算符优先关系矩阵private String[][] chars;// 存放运算符栈(index=0表示栈底，index=size()-1表示栈顶)private ArrayList<String> stack;// 存放输入串private ArrayList<String> inStr;// 存放输出串private ArrayList<String> outStr;

public static void main(String[] args) {

InversePolish inversePolish = new InversePolish();String str = "a*(b+c)/d";inversePolish.init(str);inversePolish.mainFunc();for (String currentChar : inversePolish.outStr) {System.out.print(currentChar);}}

private void mainFunc() {

// 从左往右扫描中缀表达式label1: for (String currentChar : this.inStr) {// 输入串为空?if (currentChar.equals("#")) {// 栈为空？while (true) {if (this.stack.get(this.stack.size() - 1).equals("#")) {break label1;} else {// 退栈输出this.outStr.add(this.stack.get(this.stack.size() - 1));this.stack.remove(this.stack.size() - 1);continue;}}} else {// 运算符？if (!this.isChar(currentChar)) {// 输出this.outStr.add(currentChar);continue;} else {while (true) {// 栈是否为空？if (this.stack.get(this.stack.size() - 1).equals("#")) {// 入栈this.stack.add(currentChar);continue label1;} else {// 比较当前运算符与栈顶运算符的优先级if (isFirst(currentChar)) {// 入栈this.stack.add(currentChar);continue label1;} else {// 当前运算符是')'if (currentChar.equals(")")) {while (true) {// 栈顶为'('if (this.stack.get(this.stack.size() - 1).equals("(")) {// 退栈this.stack.remove(this.stack.size() - 1);continue label1;} else {// 栈为空？if (this.stack.get(this.stack.size() - 1).equals("#")) {System.out.println("ERROR");} else {// 退栈输出this.outStr.add(this.stack.get(this.stack.size() - 1));this.stack.remove(this.stack.size() - 1);continue;}}}} else {// 退栈输出this.outStr.add(this.stack.get(this.stack.size() - 1));this.stack.remove(this.stack.size() - 1);continue;}}}}}}}}

private void init(String str) {

System.out.println(str);this.stack = new ArrayList<>();this.inStr = new ArrayList<>();this.outStr = new ArrayList<>();// 输入运算符优先关系矩阵8*7this.chars = new String[][] { { ">", ">", "<", "<", "<", "<", ">" }, { ">", ">", "<", "<", "<", "<", ">" },{ ">", ">", ">", ">", "<", "<", ">" }, { ">", ">", ">", ">", "<", "<", ">" },{ ">", ">", ">", ">", ">", "<", ">" }, { "<", "<", "<", "<", "<", "<", "=" },{ ">", ">", ">", ">", ">", "", ">" }, { "<", "<", "<", "<", "<", "<", "<" } };

// 输入输入串String[] temps = str.split("");for (String temp : temps) {this.inStr.add(temp);}this.inStr.add("#");// 输入运算符栈this.stack.add("#");}

// 运算符？private boolean isChar(String currentChar) {

return "+-*/↑()".contains(currentChar);}

// 比较当前运算符与栈顶运算符的优先级private boolean isFirst(String currentChar) {

String stackChar = this.stack.get(this.stack.size() - 1);int x = "+-*/↑()#".indexOf(stackChar);int y = "+-*/↑()#".indexOf(currentChar);return this.chars[x][y].equals("<");}
}

运行结果：

四、心得体会

实现逆波兰式的算法，难度并不大，之所以要将看似简单的中缀表达式转换为复杂的逆波兰式，原因就在于这个简单是相对人类的思维结构来说的，对计算机而言中序表达式是非常复杂的结构。相对的，逆波兰式在计算机看来却是比较简单易懂的结构。因为计算机普遍采用的内存结构是栈式结构，它执行先进后出的顺序。