Java 链接堆栈中中缀到后缀的括号检查
我有四节课Java 链接堆栈中中缀到后缀的括号检查,java,stack,Java,Stack,我有四节课 其中一个包含我的linkedstack设置 一种是infixtopostfix，用于优先级排序和转换 用于匹配的括号 评估后缀 我已经在这里设置了几乎所有内容，但不管我怎么说，它仍然返回false。 另一方面，我在上的成绩相当！stackMatch.pop（）.equals（c）不工作，因为它是带有“！”的对象类型这是个问题 我的程序简单明了： LinkedStack.java public class LinkedStack implements StackInterface {
上的成绩相当！stackMatch.pop（）.equals（c）public class LinkedStack implements StackInterface {
private Node top;
public LinkedStack() {
top = null;
} // end default constructor
public boolean isEmpty() {
return top == null;
} // end isEmpty
public void push(Object newItem) {
Node n = new Node();
n.setData(newItem);
n.setNext(top);
top = n;
} // end push
public Object pop() throws Exception {
if (!isEmpty()) {
Node temp = top;
top = top.getNext();
return temp.getData();
} else {
throw new Exception("StackException on pop: stack empty");
} // end if
} // end pop
public Object peek() throws Exception {
if (!isEmpty()) {
return top.getData();
} else {
throw new Exception("StackException on peek: stack empty");
} // end if
} // end peek
} // end LinkedStack
import java.util.*;
public class InfixToPostfix {
Parenthesis p = new Parenthesis();
LinkedStack stack = new LinkedStack();
String token = ""; // each token of the string
String output = ""; // the string holding the postfix expression
Character topOfStackObject = null; // the top object of the stack, converted to a Character Object
char charValueOfTopOfStack = ' '; // the primitive value of the Character object
/**
* Convert an infix expression to postfix. If the expression is invalid, throws an exception.
* @param s the infix expression
* @return the postfix expression as a string
* hint: StringTokenizer is very useful to this iteratively
*/
//public String convertToPostfix(String s) throws Exception {
//}
private boolean isOperand (char c){
return ((c>= '0' && c <= '9') || (c >= 'a' && c<= 'z'));
}
public void precedence(char curOp, int val) throws Exception {
while (!stack.isEmpty()) {
char topOp = (Character) stack.pop();
// charValueOfTopOfStack = topOfStackObject.charValue();
if (topOp == '(') {
stack.push(topOp);
break;
}// it's an operator
else {// precedence of new op
int prec2;
if (topOp == '+' || topOp == '-') {
prec2 = 1;
} else {
prec2 = 2;
}
if (prec2 < val) // if prec of new op less
{ // than prec of old
stack.push(topOp); // save newly-popped op
break;
} else // prec of new not less
{
output = output + topOp; // than prec of old
}
}
}
}
import java.util.*;
public class Parenthesis{
private LinkedStack stack = new LinkedStack();
private Object openBrace;
private String outputString;
/**
* Determine if the expression has matching parenthesis using a stack
*
* @param expr the expression to be evaluated
* @return returns true if the expression has matching parenthesis
*/
public boolean match(String expr) {
LinkedStack stackMatch = new LinkedStack();
for(int i=0; i < expr.length(); i++) {
char c = expr.charAt(i);
if(c == '(')
stackMatch.push(c);
else if(c == ')'){
if (stackMatch.isEmpty() || !stackMatch.pop().equals(c))
return false;
}
}
return stackMatch.isEmpty();
}
}
public class LinkedStack implements StackInterface {
private Node top;
public LinkedStack() {
top = null;
} // end default constructor
public boolean isEmpty() {
return top == null;
} // end isEmpty
public void push(Object newItem) {
Node n = new Node();
n.setData(newItem);
n.setNext(top);
top = n;
} // end push
public Object pop() throws Exception {
if (!isEmpty()) {
Node temp = top;
top = top.getNext();
return temp.getData();
} else {
throw new Exception("StackException on pop: stack empty");
} // end if
} // end pop
public Object peek() throws Exception {
if (!isEmpty()) {
return top.getData();
} else {
throw new Exception("StackException on peek: stack empty");
} // end if
} // end peek
} // end LinkedStack
import java.util.*;
public class InfixToPostfix {
Parenthesis p = new Parenthesis();
LinkedStack stack = new LinkedStack();
String token = ""; // each token of the string
String output = ""; // the string holding the postfix expression
Character topOfStackObject = null; // the top object of the stack, converted to a Character Object
char charValueOfTopOfStack = ' '; // the primitive value of the Character object
/**
* Convert an infix expression to postfix. If the expression is invalid, throws an exception.
* @param s the infix expression
* @return the postfix expression as a string
* hint: StringTokenizer is very useful to this iteratively
*/
//public String convertToPostfix(String s) throws Exception {
//}
private boolean isOperand (char c){
return ((c>= '0' && c <= '9') || (c >= 'a' && c<= 'z'));
}
public void precedence(char curOp, int val) throws Exception {
while (!stack.isEmpty()) {
char topOp = (Character) stack.pop();
// charValueOfTopOfStack = topOfStackObject.charValue();
if (topOp == '(') {
stack.push(topOp);
break;
}// it's an operator
else {// precedence of new op
int prec2;
if (topOp == '+' || topOp == '-') {
prec2 = 1;
} else {
prec2 = 2;
}
if (prec2 < val) // if prec of new op less
{ // than prec of old
stack.push(topOp); // save newly-popped op
break;
} else // prec of new not less
{
output = output + topOp; // than prec of old
}
}
}
}
import java.util.*;
public class Parenthesis{
private LinkedStack stack = new LinkedStack();
private Object openBrace;
private String outputString;
/**
* Determine if the expression has matching parenthesis using a stack
*
* @param expr the expression to be evaluated
* @return returns true if the expression has matching parenthesis
*/
public boolean match(String expr) {
LinkedStack stackMatch = new LinkedStack();
for(int i=0; i < expr.length(); i++) {
char c = expr.charAt(i);
if(c == '(')
stackMatch.push(c);
else if(c == ')'){
if (stackMatch.isEmpty() || !stackMatch.pop().equals(c))
return false;
}
}
return stackMatch.isEmpty();
}
}
import java.util.*；
公共类InfixToPostfix{
圆括号p=新圆括号（）；
LinkedStack堆栈=新LinkedStack（）；
String token=”“；//字符串的每个标记
String output=“；//包含后缀表达式的字符串
Character topOfStackObject=null；//堆栈的顶部对象，转换为Character对象
char charValueOfTopOfStack=''；//字符对象的基本值
/**
*将中缀表达式转换为后缀。如果表达式无效，则引发异常。
*@param是中缀表达式
*@以字符串形式返回后缀表达式
*提示：StringTokenizer对于这种迭代方式非常有用
*/
//公共字符串convertToPostfix（字符串s）引发异常{
//}
专用布尔等标量（字符c）{
返回（（c>='0'&&c='a'&&c问题可能是，您正在尝试测试匹配的（出现时当前位于堆栈顶部）
，但在中，c
是默认字符，）
，因此您测试）
是否位于堆栈顶部，而不是（
）