java实现逆波兰式算法（方程式运算）

感谢这位原文网友，原文来自http://blog.csdn.net/yunxiang/article/details/1918717

代码不多只有三个类，第一个是对象类，第二个是解析类（主类，外面就用这个类的接口），第三个是计算类
package expression1;

import java.util.*;

//栈类    
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();
}
}


第二个是解析类（主类，外面就用这个类的接口）

package expression1;

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, "+-*/()", true);
while (st.hasMoreElements()) {
String s = st.nextToken();
expression.add(s);
}
}

// 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(")")) {
while (true) {

if (aStack.top != -1 && !((String) aStack.top()).equals("(")) {
operator = (String) aStack.pop();
right.add(operator);
} else {
if (aStack.top != -1)
aStack.pop();
break;
}
}
} else {
while (true) {
if (aStack.top != -1
&& Calculate.priority((String) expression.get(position)) <= Calculate
.priority((String) aStack.top())) {
operator = (String) aStack.pop();
if (!operator.equals("("))
right.add(operator);
} else {
break;
}

}
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();
if (!operator.equals("("))
right.add(operator);
}
}

// 对右序表达式进行求值
private void getResult() {
this.toRight();
for (int i = 0; i < right.size(); i++) {
System.out.println(right.get(i));
}
Stacks aStack = new Stacks();
String op1, op2, is = null;
Iterator it = right.iterator();

while (it.hasNext()) {
is = (String) it.next();
if (Calculate.isOperator(is)) {
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);
}

public static void main(String avg[]) {
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!!!");
}
}
}

第三个是计算类
package expression1;

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(")"))
return true;
else
return false;
}

// 设置操作符号的优先级别
public static int priority(String operator) {
if (operator.equals("+") || operator.equals("-"))
return 1;
else if (operator.equals("*") || operator.equals("/"))
return 2;
else
return 0;
}

// 做2值之间的计算
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) {
System.out.println("input has something wrong!");
return "Error";
}
}
}