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计算器JAVA中式转逆波兰式算法

2009-05-17 19:09 225 查看
/**
* 制作人：李金泉 学校：桂林工学院 QQ：55204816 邮箱：vip_ljq@yeah.net
* 在网上面看到一些例子，运行起来会出错，所以今天来修改一下，加了一些功能；
* 希望有助于大家学习
*/
package expression;
import java.io.*;
import java.util.*;
public class Expression
{
private ArrayList expression = new ArrayList();// 存储中序表达式

private ArrayList right = new ArrayList();// 存储右序表达式

private String result;// 结果

// 依据输入信息创建对象，将数值与操作符放入ArrayList中
private Expression(String input)
{
StringTokenizer st = new StringTokenizer(input, "+-*/()SCTL", true);
while (st.hasMoreElements())
{
expression.add(st.nextToken());
}
}

// 将中序表达式转换为右序表达式
private void toRight()
{
Stacks aStack = new Stacks();
String operator;
int position = 0;
while (true)
{
if (Calculate.isOperator((String) expression.get(position)))
{
if (aStack.top == -1 || ((String) expression.get(position)).equals("("))
{
aStack.push(expression.get(position));
}
else
{
if (((String) expression.get(position)).equals(")"))
{
if (!((String) aStack.top()).equals("("))
{
operator = (String) aStack.pop();
right.add(operator);
}
//这里一定要注意了，千万不能漏，漏掉就有可能在括号的地方出错
operator= (String)aStack.pop();
//这里一定要注意了，千万不能漏，漏掉就有可能在括号的地方出错
}
else {
if (Calculate.priority((String) expression.get(position))
<= Calculate.priority((String) aStack.top())&& aStack.top != -1)
{
if (!((String) aStack.top()).equals("("))
//这个条件很重要，看到很多网上面都没有，结果求：(5*8-(6-3))*5 就会出错
//这个条件很重要，看到很多网上面都没有，结果求：(5*8-(6-3))*5 就会出错
//这个条件很重要，看到很多网上面都没有，结果求：(5*8-(6-3))*5 就会出错
{
operator = (String) aStack.pop();
right.add(operator);
}

}
aStack.push(expression.get(position));
}
}
}
else
right.add(expression.get(position));
position++;
if (position >= expression.size())
break;
}
while (aStack.top != -1) {
operator = (String) aStack.pop();
right.add(operator);
}
}

// 对右序表达式进行求值 (5*8-(6-3))*5

private void getResult()
{
this.toRight();
Stacks aStack = new Stacks();
String op1, op2, is = null;
Iterator it = right.iterator();

while (it.hasNext())
{
is = (String) it.next();
if (Calculate.isOperator(is))
{
//开始的时候没太在意这里，算三角函数一直出错，后来检查多次，终于找到这，不容易出错的让我忽略了

/*
这里是以前没改过的，超低级错误
op1 = (String) aStack.pop();
if(is.equals("S")||is.equals("C")||is.equals("T")||is.equals("L")){
aStack.push(Calculate.oneResult(is, op1));
}
op2 = (String) aStack.pop();
if(is.equals("+")||is.equals("-")||is.equals("*")||is.equals("/")){
aStack.push(Calculate.twoResult(is, op1, op2));
}
*/
if(is.equals("S")||is.equals("C")||is.equals("T")||is.equals("L")){
op1 = (String) aStack.pop();
aStack.push(Calculate.oneResult(is, op1));
}

if(is.equals("+")||is.equals("-")||is.equals("*")||is.equals("/")){
op1 = (String) aStack.pop();
op2 = (String) aStack.pop();
aStack.push(Calculate.twoResult(is, op1, op2));
}
}
else
aStack.push(is);
}
result = (String) aStack.pop();
it = expression.iterator();
while (it.hasNext())
{
System.out.print((String) it.next());
}
System.out.println("=" + result);
return ;
}

public static void main(String[] args)
{
try {
System.out.println("Input a expression:");
BufferedReader is = new BufferedReader(new InputStreamReader(System.in));
for (;;)
{
String input = new String();
input = is.readLine().trim();
if (input.equals("q"))
break;
else
{
Expression boya = new Expression(input);
boya.getResult();
}
System.out.println("Input another expression or input 'q' to quit:");
}
is.close();
}
catch (IOException e)
{
System.out.println("Wrong input!!!");
}

}
}

package expression;
import javax.swing.JOptionPane;

public class Calculate {
// 判断是否为操作符号
public static boolean isOperator(String operator) {
if (operator.equals("+") || operator.equals("-")
|| operator.equals("*") || operator.equals("/")
|| operator.equals("(") || operator.equals(")")
|| operator.equals("T") || operator.equals("S")
|| operator.equals("C") || operator.equals("L"))
return true;
else
return false;
}

// 设置操作符号的优先级别
//这里要记住，操作数越少，优先级就越高，一点小心得，给大家分享一下吧
//还有，这里的“）”优先级是最低的，虽然没写明，但是能看的出来
public static int priority(String operator)
{
if (operator.equals("+") || operator.equals("-")
|| operator.equals("("))
return 1;
else if (operator.equals("*") || operator.equals("/"))
return 2;
else if (operator.equals("S") || operator.equals("C")||operator.equals("T") || operator.equals("L"))
return 3;
else
return 0;
}

// 有两个操作数的函数运算
public static String twoResult(String operator, String a, String b) {
try {
String op = operator;
String rs = new String();
double x = Double.parseDouble(b);
double y = Double.parseDouble(a);
double z = 0;
if (op.equals("+"))
z = x + y;
else if (op.equals("-"))
z = x - y;
else if (op.equals("*"))
z = x * y;
else if (op.equals("/"))
z = x / y;
else
z = 0;
return rs + z;
} catch (NumberFormatException e) {
JOptionPane.showMessageDialog( null, "input has something wrong!","错误提示",JOptionPane.WARNING_MESSAGE);
return "Error";
}
}
// 有一个操作数的函数运算
public static String oneResult(String operator, String a) {
try {
String op = operator;
String rs = new String();
double x = Double.parseDouble(a);
double z = 0;
if (op.equals("S"))
z = Math.sin(x);
else if (op.equals("C"))
z = Math.cos(x) ;
else if (op.equals("T"))
z =Math.tan(x) ;
else if (op.equals("L"))
z = Math.log(x);
else
z = 0;
return rs + z;
} catch (NumberFormatException e) {
JOptionPane.showMessageDialog( null, "input has something wrong!","错误提示",JOptionPane.WARNING_MESSAGE);

return "Error";
}
}

}
package expression;
import java.util.*;
public class Stacks {
private LinkedList list=new LinkedList();
int top=-1;
public void push(Object value){
top++;
list.addFirst(value);
}
public Object pop(){
Object temp=list.getFirst();
top--;
list.removeFirst();
return temp;

}
public Object top(){
return list.getFirst();
}
}
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