我应该写一个程序将中缀转换为postfix。它适用于某些人,但其他时候不正确。特别是在包含parantheses的中缀表达式上。任何人都可以告诉我为什么这是错的?例如,中缀表达式
( ( 5 + 5 * ( 6 - 2 ) + 4 ^ 2 ) * 8 )
返回5562-*42^++8*((2
。
import java.io.*;
import java.util.Scanner;
public class InfixToPostfix
{
//class attributes
private char curValue;
private String postfix;
private LineWriter lw;
private ObjectStack os;
//constructor
public InfixToPostfix(LineWriter l, ObjectStack o)
{
curValue = ' ';
lw=l;
os=o;
}
public String conversion(String buf)
{
String temp =" ";
StringBuffer postfixStrBuf= new StringBuffer(temp);
char popped= new Character(' ');
char topped=' ';
for (int i=0; i<buf.length(); i++)
{
curValue= buf.charAt(i);
if (curValue == '(')
os.push(curValue);
if (curValue == ')')
{
while (popped != '(')
{
popped = ((Character)os.pop());
if (popped != '(')
postfixStrBuf.append(popped);
}
}
if (isOperator(curValue))
{
if( os.isEmpty())
os.push((Character)(curValue));
else
topped=((Character)os.top());
if ( (priority(topped)) >= (priority(curValue)) && (topped != ' ') )
{
popped = ((Character)os.pop());
if (popped != '(')
postfixStrBuf.append(popped);
//if it is a left paranthess, we want to go ahead and push it anyways
os.push((Character)(curValue));
}
if ( (priority(topped)) < (priority(curValue)) && (topped != ' ') )
os.push((Character)(curValue));
}
else if (!isOperator(curValue) && (curValue != ' ') && (curValue != '(' ) && (curValue != ')' ))
postfixStrBuf.append(curValue);
}
//before you grab the next line of the file , pop off whatever is remaining off the stack and append it to
//the infix expression
getRemainingOp(postfixStrBuf);
return postfix;
//postfixStrBuf.delete(0, postfixStrBuf.length());
}
public int priority(char curValue)
{
switch (curValue)
{
case '^': return 3;
case '*':
case '/': return 2;
case '+':
case '-': return 1;
default : return 0;
}
}
public boolean isOperator(char curValue)
{
boolean operator = false;
if ( (curValue == '^' ) || (curValue == '*') || (curValue == '/') || (curValue == '+' ) || (curValue == '-') )
operator = true;
return operator;
}
public String getRemainingOp(StringBuffer postfixStrBuf)
{
char popped=' ';
while ( !(os.isEmpty()) )
{
opped = ((Character)os.pop());
postfixStrBuf.append(popped);
}
postfix=postfixStrBuf.toString();
return postfix;
}
}
答案 0 :(得分:0)
我只会发布内循环应该是什么样子(到处都没有铸件):
if (curValue == '(') {
os.push(curValue);
} else if (curValue == ')') {
if(!os.isEmpty()) {
topped = os.pop();
while (!os.isEmpty() && (topped != '(')) {
postfixStrBuf.append(topped);
topped = os.pop();
}
}
} else if (isOperator(curValue)) {
if (os.isEmpty()) {
os.push(curValue);
} else {
while(!os.isEmpty() && (priority(os.top()) >= priority(curValue))) {
popped = os.pop();
postfixStrBuf.append(popped);
}
os.push(curValue);
}
} else if (curValue != ' ') {
postfixStrBuf.append(curValue);
}
披露:已经很晚了所以我希望它没事。您应该修改变量的初始化方式并返回getRemainingOp
方法。