java - 不使用 "clone"复制堆栈或队列

Question

不使用克隆复制堆栈和队列。例如，当我调用传递堆栈的方法时，我无法修改保留传递的原始堆栈。我需要复制/克隆传递的堆栈以在方法中更改/使用。

我只能使用Stack.java(附件)。我创建了以下辅助方法:

public static Stack<CalendarDate> qToS(Queue<CalendarDate> q) {
    Stack<CalendarDate> s = new Stack<CalendarDate>();
    while (!q.isEmpty()) {
       CalendarDate n = q.remove();
       s.push(n);
    }
    return s; // Return stack s
}

public static Queue<CalendarDate> sToQ(Stack<CalendarDate> s) {
    Queue<CalendarDate> q = new LinkedList<CalendarDate>();
    while (!s.empty()) {
       CalendarDate n = s.pop();
       q.add(n);
    }
    return q; // Return queue q
}

/*      
    Provided as a Stack Class alternative
    Limits user to actual Stack methods
    so Vector<E> is not available
*/
public class Stack<E> {
    // avoid blanked import of java.util
    private java.util.Stack<E> secret;

    // default constructor
    public Stack() {
        secret = new java.util.Stack<E>();
    } 

    // empty that collection
    public void clear() {
        secret.clear();
    }

    // should be order constant
    public int size() {
        return secret.size();
    }

    // simply have push call push from API
    public E push(E a) {
        secret.push(a);
        return a;
    }

    // And, empty calls empty from API
    public boolean empty() {
        return secret.empty();
    }

    // And my pop() uses pop() form JAVA API
    public E pop() {
        return secret.pop();
    }

    // My peek uses their peek
    public E peek() {
        return secret.peek();
    }

    // Following are not basic Stack operations
    // but needed to do some simple testing

    // toString is probably not O(constant)
    public String toString() {
        return secret.toString();
    }

}

我的解决方案

public static Stack<CalendarDate> sToS(Stack<CalendarDate> orgin) {
        // Create a temp stack
        Stack<CalendarDate> temp = new Stack<CalendarDate>();

        // Move all values from origin
        // stack to temp stack using pop and push
        while (!orgin.empty()) {
            CalendarDate n = orgin.pop();
            temp.push(n);    // push here for the same order
        }

        // Create a copy stack
        Stack<CalendarDate> copy = new Stack<CalendarDate>();

        // Move all values from temp stack to
        // both origin and copy stacks at the same time
        while (!temp.empty()) {
            CalendarDate n = temp.pop();
            copy.push(n);    // push here for the same order
            orgin.push(n);  
        }

        return copy;
    }

Answer 1

想象一个场景，您有三个堆栈:堆栈 A(要从中复制的堆栈)、堆栈 B(要复制到的目标)和堆栈临时(辅助堆栈)。

Step 1: (The Initial Stack)

|1|  | |  | |
|2|  | |  | |
|3|  | |  | |

 A   TEMP  B

Step 2: (Move elements from Stack A to Temp Stack)

| |  | |  | |
|2|  | |  | |
|3|  |1|  | |

 A   TEMP  B

| |  | |  | |
| |  |2|  | |
|3|  |1|  | |

 A   TEMP  B

| |  |3|  | |
| |  |2|  | |
| |  |1|  | |

 A   TEMP  B

Step 3: (Move elements from Temp stack to Stack A & B)

| |  | |  | |
| |  |2|  | |
|3|  |1|  |3|

 A   TEMP  B

| |  | |  | |
|2|  | |  |2|
|3|  |1|  |3|

 A   TEMP  B

|1|  | |  |1|
|2|  | |  |2|
|3|  | |  |3|

 A   TEMP  B

充分理解的最好方法是举个例子并亲自尝试一下。或者您可以简单地在编码平台上搜索方法，例如 GeeksForGeeks