This question already has answers here:
Handling parenthesis while converting infix expressions to postfix expressions
(2个答案)
3年前关闭。
我必须制作一个程序,将以Infix表示法编写的表达式更改为Postfix表示法。我开始使用括号时遇到问题。例如,当我输入“ a +(c-h)/(b * d)”时,应显示为“ ac + h-b / d *”,而当它应显示为“ a c h-b d * / +”时。非常感谢您的帮助。谢谢。
(2个答案)
3年前关闭。
我必须制作一个程序,将以Infix表示法编写的表达式更改为Postfix表示法。我开始使用括号时遇到问题。例如,当我输入“ a +(c-h)/(b * d)”时,应显示为“ ac + h-b / d *”,而当它应显示为“ a c h-b d * / +”时。非常感谢您的帮助。谢谢。
import java.util.Scanner;
import java.util.Stack;
public class PostfixConverter {
static private String expression;
private Stack<Character> stack = new Stack<Character>();
public PostfixConverter(String infixExpression) {
expression = infixExpression;
}
public String infixToPostfix() {
String postfixString = "";
for (int index = 0; index < expression.length(); ++index) {
char value = expression.charAt(index);
if (value == '(') {
} else if (value == ')') {
Character oper = stack.peek();
while (!(oper.equals('(')) && !(stack.isEmpty())) {
stack.pop();
postfixString += oper.charValue();
}
} else if (value == '+' || value == '-') {
if (stack.isEmpty()) {
stack.push(value);
} else {
Character oper = stack.peek();
while (!(stack.isEmpty() || oper.equals(('(')) || oper.equals((')')))) {
stack.pop();
postfixString += oper.charValue();
}
stack.push(value);
}
} else if (value == '*' || value == '/') {
if (stack.isEmpty()) {
stack.push(value);
} else {
Character oper = stack.peek();
while (!oper.equals(('+')) && !oper.equals(('-')) && !stack.isEmpty()) {
stack.pop();
postfixString += oper.charValue();
}
stack.push(value);
}
} else {
postfixString += value;
}
}
while (!stack.isEmpty()) {
Character oper = stack.peek();
if (!oper.equals(('('))) {
stack.pop();
postfixString += oper.charValue();
}
}
return postfixString;
}
public static void main(String[] args) {
System.out.println("Type an expression written in Infix notation: ");
Scanner input = new Scanner(System.in);
String expression = input.next();
PostfixConverter convert = new PostfixConverter(expression);
System.out.println("This expression writtien in Postfix notation is: \n" + convert.infixToPostfix());
}
}
最佳答案
您提供的代码类似于this
。但是该代码也不起作用。
我已经更新了您的代码和更改的added the comments
。
import java.util.Scanner;
import java.util.Stack;
public class PostfixConverter {
static private String expression;
private Stack<Character> stack = new Stack<Character>();
public PostfixConverter(String infixExpression) {
expression = infixExpression;
}
public String infixToPostfix() {
String postfixString = "";
for (int index = 0; index < expression.length(); ++index) {
char value = expression.charAt(index);
if (value == '(') {
stack.push('('); // Code Added
} else if (value == ')') {
Character oper = stack.peek();
while (!(oper.equals('(')) && !(stack.isEmpty())) {
stack.pop();
postfixString += oper.charValue();
if (!stack.isEmpty()) // Code Added
oper = stack.peek(); // Code Added
}
stack.pop(); // Code Added
} else if (value == '+' || value == '-') {
if (stack.isEmpty()) {
stack.push(value);
} else {
Character oper = stack.peek();
while (!(stack.isEmpty() || oper.equals(('(')) || oper.equals((')')))) {
oper = stack.pop(); // Code Updated
postfixString += oper.charValue();
}
stack.push(value);
}
} else if (value == '*' || value == '/') {
if (stack.isEmpty()) {
stack.push(value);
} else {
Character oper = stack.peek();
// while condition updated
while (!oper.equals(('(')) && !oper.equals(('+')) && !oper.equals(('-')) && !stack.isEmpty()) {
oper = stack.pop(); // Code Updated
postfixString += oper.charValue();
}
stack.push(value);
}
} else {
postfixString += value;
}
}
while (!stack.isEmpty()) {
Character oper = stack.peek();
if (!oper.equals(('('))) {
stack.pop();
postfixString += oper.charValue();
}
}
return postfixString;
}
public static void main(String[] args) {
System.out.println("Type an expression written in Infix notation: ");
Scanner input = new Scanner(System.in);
String expression = input.next();
PostfixConverter convert = new PostfixConverter(expression);
System.out.println("This expression writtien in Postfix notation is: \n" + convert.infixToPostfix());
}
}
10-04 17:26