Question

public class Matcher {

public MatchRtnCode matcher(String str){

    Stack<Character> s=new LLStack<Character>();
    for(int i =0; i<str.length(); i++){
        char c=str.charAt(i);
        if(c=='{' || c=='[' || c=='('){
            s.push(c);
        }else if (c=='}'){
            Character temp=s.pop();
            if(temp!='{'){
                s.push(temp);
                return MatchRtnCode.UNEXPECTED_SYMBOL;
            }


        }else if (c==']'){
            Character temp=s.pop();
            if(temp!='['){
                s.push(temp);
                return MatchRtnCode.UNEXPECTED_SYMBOL;
            }

        }else if (c==')'){
            Character temp=s.pop();
            if(temp!='('){
                s.push(temp);
                return MatchRtnCode.UNEXPECTED_SYMBOL;
            }

        }else{

        }
    }
    Character temp = s.pop();
    if(temp=='{' || temp=='[' || temp=='('){
        return MatchRtnCode.TOO_MANY_OPEN_SYMBOLS;
    }else if(temp=='}' || temp==']' || temp==')'){
        return MatchRtnCode.TOO_MANY_CLOSED_SYMBOLS;
    }else {
        return MatchRtnCode.GOOD_STRING;
    }

}
}

Stack类是：

 import java.util.EmptyStackException;

 public interface Stack<E> {

public void push(E data) throws StackFullException;
public E pop() throws EmptyStackException;
public boolean isEmpty() throws EmptyStackException;

}

LLStack类是：

 import java.util.EmptyStackException;


public class LLStack<E> implements Stack<E> {
public Node top;
private class Node{
    public E data;
    public Node next;
    public Node(E data, Node next){
        this.data=data;
        this.next=next;
    }
}

public LLStack() {
    top = null;
}

@Override
public void push(E data) throws StackFullException {

    top=new Node(data, top);
}

@Override
public E pop() throws EmptyStackException {
    if(top==null){
        throw new EmptyStackException();
    }
    E data=top.data;
    top=top.next;
    return data;
}

@Override
public boolean isEmpty() {
    return top==null;
}
}

我正在检查一个字符串示例：{[（ab）]}以确保{[（有）}的数量相同。现在它只是打印出EmptyStackError。我该如何解决这个问题？

Answer 1

 if(s.pop()=='{'){
     s.pop();
 }

pop（）改变堆栈的状态。一旦你弹出（）并且不需要它，你需要偷看或记得重新开启。

所以试试：

 char temp = s.pop();
 if(temp != '{') {
     s.push(temp);
 }

或者只是偷看方法：

public char peek() {
    char c = s.pop();
    s.push(c);
    return c;
}

使用此代替构造if (s.pop() == '}')

使用链接列表堆栈

1 个答案: