使用 java 中的方法交換單鏈表中的兩個節點
[英]swap two nodes in singly linked lists using method in java
問題描述
package linkedlists;
public class SinglyLinkedList<E> implements Cloneable {
//---------------- nested Node class ----------------
/**
* Node of a singly linked list, which stores a reference to its
* element and to the subsequent node in the list (or null if this
* is the last node).
*/
private static class Node<E> {
/** The element stored at this node */
private E element; // reference to the element stored at this node
/** A reference to the subsequent node in the list */
private Node<E> next; // reference to the subsequent node in the list
/**
* Creates a node with the given element and next node.
*
* @param e the element to be stored
* @param n reference to a node that should follow the new node
*/
public Node(E e, Node<E> n) {
element = e;
next = n;
}
// Accessor methods
/**
* Returns the element stored at the node.
* @return the element stored at the node
*/
public E getElement() { return element; }
/**
* Returns the node that follows this one (or null if no such node).
* @return the following node
*/
public Node<E> getNext() { return next; }
// Modifier methods
/**
* Sets the node's next reference to point to Node n.
* @param n the node that should follow this one
*/
public void setNext(Node<E> n) { next = n; }
} //----------- end of nested Node class -----------
// instance variables of the SinglyLinkedList
/** The head node of the list */
private Node<E> head = null; // head node of the list (or null if empty)
/** The last node of the list */
private Node<E> tail = null; // last node of the list (or null if empty)
/** Number of nodes in the list */
private int size = 0; // number of nodes in the list
/** Constructs an initially empty list. */
public SinglyLinkedList() { } // constructs an initially empty list
// access methods
/**
* Returns the number of elements in the linked list.
* @return number of elements in the linked list
*/
public int size() { return size; }
/**
* Tests whether the linked list is empty.
* @return true if the linked list is empty, false otherwise
*/
public boolean isEmpty() { return size == 0; }
/**
* Returns (but does not remove) the first element of the list
* @return element at the front of the list (or null if empty)
*/
public E first() { // returns (but does not remove) the first element
if (isEmpty()) return null;
return head.getElement();
}
/**
* Returns (but does not remove) the last element of the list.
* @return element at the end of the list (or null if empty)
*/
public E last() { // returns (but does not remove) the last element
if (isEmpty()) return null;
return tail.getElement();
}
// update methods
/**
* Adds an element to the front of the list.
* @param e the new element to add
*/
public void addFirst(E e) { // adds element e to the front of the list
head = new Node<>(e, head); // create and link a new node
if (size == 0)
tail = head; // special case: new node becomes tail also
size++;
}
/**
* Adds an element to the end of the list.
* @param e the new element to add
*/
public void addLast(E e) { // adds element e to the end of the list
Node<E> newest = new Node<>(e, null); // node will eventually be the tail
if (isEmpty())
head = newest; // special case: previously empty list
else
tail.setNext(newest); // new node after existing tail
tail = newest; // new node becomes the tail
size++;
}
/**
* Removes and returns the first element of the list.
* @return the removed element (or null if empty)
*/
public E removeFirst() { // removes and returns the first element
if (isEmpty()) return null; // nothing to remove
E answer = head.getElement();
head = head.getNext(); // will become null if list had only one node
size--;
if (size == 0)
tail = null; // special case as list is now empty
return answer;
}
@SuppressWarnings({"unchecked"})
public boolean equals(Object o) {
if (o == null) return false;
if (getClass() != o.getClass()) return false;
SinglyLinkedList other = (SinglyLinkedList) o; // use nonparameterized type
if (size != other.size) return false;
Node walkA = head; // traverse the primary list
Node walkB = other.head; // traverse the secondary list
while (walkA != null) {
if (!walkA.getElement().equals(walkB.getElement())) return false; //mismatch
walkA = walkA.getNext();
walkB = walkB.getNext();
}
return true; // if we reach this, everything matched successfully
}
@SuppressWarnings({"unchecked"})
public SinglyLinkedList<E> clone() throws CloneNotSupportedException {
// always use inherited Object.clone() to create the initial copy
SinglyLinkedList<E> other = (SinglyLinkedList<E>) super.clone(); // safe cast
if (size > 0) { // we need independent chain of nodes
other.head = new Node<>(head.getElement(), null);
Node<E> walk = head.getNext(); // walk through remainder of original list
Node<E> otherTail = other.head; // remember most recently created node
while (walk != null) { // make a new node storing same element
Node<E> newest = new Node<>(walk.getElement(), null);
otherTail.setNext(newest); // link previous node to this one
otherTail = newest;
walk = walk.getNext();
}
}
return other;
}
public int hashCode() {
int h = 0;
for (Node walk=head; walk != null; walk = walk.getNext()) {
h ^= walk.getElement().hashCode(); // bitwise exclusive-or with element's code
h = (h << 5) | (h >>> 27); // 5-bit cyclic shift of composite code
}
return h;
}
/**
* Produces a string representation of the contents of the list.
* This exists for debugging purposes only.
*/
public String toString() {
StringBuilder sb = new StringBuilder("(");
Node<E> walk = head;
while (walk != null) {
sb.append(walk.getElement());
if (walk != tail)
sb.append(", ");
walk = walk.getNext();
}
sb.append(")");
return sb.toString();
}
public void swapNodes(Node<E> num1, Node<E> num2) {
Node<E> num1Prev = this.head;
Node<E> num2Prev = this.head;
if (num1 == num2 )
return ;
while((num1Prev != null)&&(num1Prev.getNext() != num1)){
num1Prev = num1Prev.getNext();
}
while((num2Prev != null)&&(num2Prev.getNext() != num2)){
num2Prev = num2Prev.getNext();
}
if(num2Prev == num1) {
num1.setNext(num2.getNext());
num2.setNext(num1);
num1Prev.setNext(num2);
}
else if(num1Prev == num2) {
num2.setNext(num1.getNext());
num1.setNext(num2);
num2Prev.setNext(num1);
}
else {
Node<E> tmp = num1.getNext();
num1.setNext(num2.getNext());
num2.setNext(tmp);
num1Prev.setNext(num2);
num2Prev.setNext(num1);
}
}
//main method
public static void main(String[] args)
{
SinglyLinkedList<String> list = new SinglyLinkedList<String>();
list.addFirst("MSP");
list.addLast("ATL");
list.addLast("BOS");
//
list.addFirst("LAX");
System.out.println(list);
//
SinglyLinkedList<String> swap = new SinglyLinkedList<String>();
swap.addFirst("1");
swap.addLast("2");
swap.addLast("3");
swap.addLast("4");
swap.addLast("5");
System.out.println("Original list: " + swap);
swap.swapNodes("2","5");
System.out.println("After Swapping list: " + swap);
}
}
任務：在本練習中，您將從第 2 周的講座示例中將方法 swapNodes 添加到 SinglyLinkedList class。 這個方法應該交換兩個節點 node1 和 node2（而不僅僅是它們的內容），因為它只引用了 node1 和 node2。 新方法應該檢查node1和node2是否是同一個節點等。編寫main方法來測試swapNodes方法。 提示：您可能需要遍歷列表。
我做了這個方法
public void swapNodes(Node<E> num1, Node<E> num2) {
Node<E> num1Prev = this.head;
Node<E> num2Prev = this.head;
if (num1 == num2 )
return ;
while((num1Prev != null)&&(num1Prev.getNext() != num1)){
num1Prev = num1Prev.getNext();
}
while((num2Prev != null)&&(num2Prev.getNext() != num2)){
num2Prev = num2Prev.getNext();
}
if(num2Prev == num1) {
num1.setNext(num2.getNext());
num2.setNext(num1);
num1Prev.setNext(num2);
}
else if(num1Prev == num2) {
num2.setNext(num1.getNext());
num1.setNext(num2);
num2Prev.setNext(num1);
}
else {
Node<E> tmp = num1.getNext();
num1.setNext(num2.getNext());
num2.setNext(tmp);
num1Prev.setNext(num2);
num2Prev.setNext(num1);
}
}
然后，創建了一個實例來檢查它是否工作，但它在這里向我顯示一個錯誤swap.swapNodes("2", "5"); 有誰知道是什么問題？ 謝謝
1 個解決方案
解決方案1
0 2021-02-02 21:28:10
它拋出錯誤，因為在您的swapNodes function 中，您希望傳遞兩個類型為Node<E>的參數，但您傳遞的是E (String) 。 因此，您必須更改 function 的簽名或傳遞您添加到列表中的Node<E> 。
以下是使用E類型參數執行swapNodes function 的方法：
public void swapNodes(E element1, E element2) {
if (element1 == element2 || element1 == null || element2 == null || isEmpty()) return;
Node<E> num1 = findNode(element1);
Node<E> num2 = findNode(element2);
if (num1 == null || num2 == null) return;
Node<E> num1Prev = this.head;
Node<E> num2Prev = this.head;
while(num1Prev != null && num1Prev.getNext() != num1){
num1Prev = num1Prev.getNext();
}
while(num2Prev != null && num2Prev.getNext() != num2){
num2Prev = num2Prev.getNext();
}
if (num1Prev.getNext() == null || num2Prev.getNext() == null) return;
if (num2Prev.equals(num1)) {
num1.setNext(num2.getNext());
num2.setNext(num1);
num1Prev.setNext(num2;
} else if (num1Prev.equals(num2)) {
num2.setNext(num1.getNext());
num1.setNext(num2);
num2Prev.setNext(num1);
} else {
Node<E> tmp = num1.getNext();
num1.setNext(num2.getNext());
num2.setNext(tmp);
num1Prev.setNext(num2);
num2Prev.setNext(num1);
}
}
findNode function 可能如下所示：
public Node<E> findNode(E element) {
if (isEmpty()) return null;
Node<E> node = this.head;
while(node != null) {
if (node.getElement() == element) return node;
}
return null;
}
通過在Java中從每個元素獲取2個元素來合並兩個單鏈列表
[英]Merge two singly linked lists by getting 2 elements from each in Java
使用隊列創建實現為Java中單鏈列表樹的文件系統
[英]Creating a filesystem implemented as a tree of singly linked lists in Java, using queues
聲明:本站的技術帖子網頁，遵循CC BY-SA 4.0協議，如果您需要轉載，請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.