大家好我是java的新手,所以我真的非常感谢任何帮助。 好的,这就是问题所在: 我有一个列表类和一个listNode类, 列表Class由名称,firstNode和lastNode表示。 firstNode和lastNode来自listNode类型, listNode由Object(例如data或Object o)和nextNode表示,nextNode指向列表中的下一个Node,它也来自listNode类型。
列表类:
public class List {
private ListNode firstNode;
private ListNode lastNode;
private String name;
public List() {
this("list");
}
public List(String listName) {
name = listName;
firstNode = lastNode = null;
}
public void insertAtFront(Object insertItem) {
if (isEmpty())
firstNode = lastNode = new ListNode(insertItem);
else
firstNode = new ListNode(insertItem, firstNode);
}
public void insertAtBack(Object insertItem) {
if (isEmpty())
firstNode = lastNode = new ListNode(insertItem);
else
lastNode = lastNode.nextNode = new ListNode(insertItem);
}
public Object removeFromFront() throws EmptyListException {
if (isEmpty())
throw new EmptyListException(name);
Object removedItem = firstNode.data;
if (firstNode == lastNode)
firstNode = lastNode = null;
else
firstNode = firstNode.nextNode;
return removedItem;
}
public Object removeFromBack() throws EmptyListException {
if (isEmpty())
throw new EmptyListException(name);
Object removedItem = lastNode.data;
if (firstNode == lastNode)
firstNode = lastNode = null;
else {
ListNode current = firstNode;
while (current.nextNode != lastNode)
current = current.nextNode;
lastNode = current;
current.nextNode = null;
}
return removedItem;
}
public boolean isEmpty() {
return firstNode == null;
}
public void print() {
if (isEmpty()) {
System.out.printf("Empty %s\n", name);
return;
}
System.out.printf("The %s is : ", name);
ListNode current = firstNode;
while (current != null) {
System.out.printf("%s", current.data);
current = current.nextNode;
}
System.out.println("\n");
}
@Override
public String toString() {
String stk = "(";
if(isEmpty())return "Empty List";
ListNode checkNode = firstNode;
while (checkNode != null) {
stk += checkNode.data.toString()+ " , ";
checkNode = checkNode.nextNode;
}
return stk+")";
}
public ListNode removeAt (int k){
if(k<=0 || k>getLength())
try{
throw new IllegalValues();
}catch(IllegalValues iv){
iv.printStackTrace();
return null;
}
ListNode newNode = firstNode;
if(k==1){
newNode = firstNode;
firstNode = firstNode.nextNode;
return newNode;
}
if(k==2){
newNode = firstNode.nextNode;
firstNode.nextNode = firstNode.nextNode.nextNode;
return newNode;
}
if(k==3){
newNode = firstNode.nextNode;
firstNode.nextNode.nextNode = firstNode.nextNode.nextNode.nextNode;
return newNode;
}
if(k==4){
newNode = firstNode.nextNode;
firstNode.nextNode.nextNode.nextNode = firstNode.nextNode.nextNode.nextNode.nextNode;
return newNode;
}
return newNode;
}
public int getLength(){
ListNode checkNode = firstNode;
int count =0;
while (checkNode != null) {
count++;
checkNode = checkNode.nextNode;
}
return count;
}
}
listNode class:
public class ListNode {
Object data;
ListNode nextNode;
public ListNode(Object o) {
this(o, null);
}
public ListNode(Object o, ListNode node) {
data = o;
nextNode = node;
}
public Object getObject() {
return data;
}
public ListNode getNext(){
return nextNode;
}
}
所以这些是我正在使用的两个类。 我的问题在于removeAt()方法我不知道如何概括它并为所有代码做出一般的答案(比如for语句)我只能通过编写每个案例来使其工作在if语句中分开。我需要编写一个for循环,以某种方式循环抛出hasNext()方法。 有任何想法吗?? 提前谢谢
答案 0 :(得分:0)
好吧,为了删除k
第n个节点(当k> 1时),总会有一个节点应该执行:
someNode.nextNode = someNode.nextNode.nextNode;
这是您要删除的节点之前的节点,这使其成为k-1
第3个节点。
您可以找到带有for循环的节点:
if (k==1) {
ListNode removedNode = firstNode;
firstNode = firstNode.nextNode;
return removedNode;
}
ListNode someNode = firstNode;
for (int i = 1; i < k - 1; i++) {
someNode = someNode.nextNode;
}
ListNode removedNode = someNode.nextNode;
someNode.nextNode = someNode.nextNode.nextNode;
return removedNode;
请注意,k == 1的情况是单独处理的。
答案 1 :(得分:0)
这是一个使用0作为第一个索引的变体:
public ListNode removeAt (int k)
{
if (k < 0 || k >= getLength())
{
try
{
throw new IllegalValues();
}
catch(IllegalValues iv)
{
iv.printStackTrace();
return null;
}
if (k == 0)
{
// first element should be removed
removeFromFront();
}
else if (k == getLength() – 1)
{
// last element should be removed
removeFromBack();
}
else
{
ListNode predecessor = firstNode;
for (int i = 0; i < k - 1; ++i)
{
predecessor = predecessor.nextNode;
}
ListNode removed = predecessor.nextNode;
predecessor.nextNode = predecessor.nextNode.nextNode;
return removed;
}