【发布时间】:2018-08-20 00:03:23
【问题描述】:
目标是通过构造函数传递数据结构(队列),并在通过方法后返回一个新队列。我创建了一个 Queue 类型的方法,可以将中缀转换为后缀顺序。问题是,当我通过构造函数传递队列时,我输出的是所有的 'a' 而不是方程本身。所以,我知道链表传递的是队列的长度,而不是字符本身。 输出:
a+b+c/(d+f)
aaaaaaaaaaa
主类:
import java.io.*;
import java.lang.*;
class Convert
{
static int Prec(char ch)
{
switch (ch)
{
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '^':
return 3;
}
return -1;
}
public static QueueADT infixToPostFix(QueueADT in)
{
QueueADT infix = in;
QueueADT result = new QueueADT();
StackADT stack = new StackADT();
while(infix.empty() == false)
{
char c = infix.dequeue();
if (Character.isLetterOrDigit(c))
result.enqueue(c);
else if (c == '(')
stack.push(c);
else if (c == ')')
{
while (!stack.empty() && stack.peek() != '(')
result.enqueue(stack.pop());
stack.pop();
}
else // an operator is encountered
{
while (!stack.empty() && Prec(c) <= Prec(stack.peek()))
result.enqueue(stack.pop());
stack.push(c);
}
}
// pop all the operators from the stack
while (!stack.empty())
result.enqueue(stack.pop());
return result;
}
public static void main(String[] args)
{
QueueADT infix = new QueueADT();
String str = "a+b+c/(d+f)";
for(int i=0; i < str.length(); i++)
{
infix.enqueue(str.charAt(i));
System.out.print(str.charAt(i));
}
QueueADT postfix = infixToPostFix(infix);
System.out.println();
while(!postfix.empty())
{
System.out.print(postfix.dequeue());
}
}
}
队列类:
public class QueueADT
{
private int size;
private Node front;
private Node rear;
public QueueADT()
{
size = 0;
front = null;
rear = null;
}
public boolean empty()
{
return(size == 0);
}
public int size()
{
return size;
}
public void enqueue(char character)
{
Node newNode = new Node();
newNode.setData(character);
newNode.setNext(null);
if(this.empty())
{
front = newNode;
}
else
rear.setNext(newNode);
rear = newNode;
size++;
}
public char dequeue()
{
char i;
i = front.getData();
size--;
if(this.empty())
rear = null;
return i;
}
public char front()
{
return front.getData();
}
}
堆栈类:
public class StackADT
{
private Node top;
private int size;
public StackADT()
{
top = null;
size = 0;
}
public boolean empty()
{
return (top == null);
}
public char peek()
{
return top.getData();
}
public int size()
{
return size;
}
public void push(char character)
{
Node newNode = new Node();
newNode.setData(character);
newNode.setNext(top);
top = newNode;
size++;
}
public char pop()
{
char i;
i = top.getData();
top = top.getNext();
size--;
return i;
}
public int onTop()
{
char i = pop();
push(i);
return i;
}
}
节点类:
public class Node
{
private char data;
private Node next;
public Node()
{
data = 0;
next = null;
}
public Node(char d)
{
data = d;
}
public Node(char d, Node n)
{
data = d;
next = n;
}
public void setData(char newData)
{
data = newData;
}
public void setNext(Node newNext)
{
next = newNext;
}
public char getData()
{
return data;
}
public Node getNext()
{
return next;
}
public void displayNode()
{
System.out.print(data);
}
}
【问题讨论】: