这个自定义类模仿了Java的LinkedList类的功能,只是它只接受整数,显然缺乏大部分功能。对于这个方法removeAll(),我将遍历列表的每个节点,并删除具有该值的所有节点。我的问题是,当列表中的第一个节点包含要删除的值时,它会忽略包含该值的所有后续节点。有什么问题吗?我是否以错误的方式移除前节点?例如,[1]->[1]->[1]应该返回一个空列表,但它离开前节点,我得到[1]
编辑:它似乎无法删除第二个节点而不是第一个节点。
这是一个类(将ListNodes存储为列表):
public class LinkedIntList {
private ListNode front; // first value in the list
// post: constructs an empty list
public LinkedIntList() {
front = null;
}
// post: removes all occurrences of a particular value
public void removeAll(int value) {
ListNode current = front; // primes loop
if (current == null) { // If empty list
return;
}
if (front.data == value) { // If match on first elem
front = current.next;
current = current.next;
}
while (current.next != null) { // If next node exists
if (current.next.data == value) { // If match at next value
current.next = current.next.next;
} else { // If not a match
current = current.next; // increment to next
}
}
}
// post: appends the given value to the end of the list
public void add(int value) {
if (front == null) {
front = new ListNode(value);
} else {
ListNode current = front;
while (current.next != null) {
current = current.next;
}
current.next = new ListNode(value);
}
}
// Sets a particular index w/ a given value
public void set(int index, int value) {
ListNode current = front;
for (int i = 0; i < index; i++) {
current = current.next;
}
current.data = value;
}
}
下面是ListNode类(负责单个"节点"):
//ListNode is a class for storing a single node of a linked
//list. This node class is for a list of integer values.
public class ListNode {
public int data; // data stored in this node
public ListNode next; // link to next node in the list
// post: constructs a node with data 0 and null link
public ListNode() {
this(0, null);
}
// post: constructs a node with given data and null link
public ListNode(int data) {
this(data, null);
}
// post: constructs a node with given data and given link
public ListNode(int data, ListNode next) {
this.data = data;
this.next = next;
}
}
实际上留在列表中的[1]元素是第二个元素,它在代码中成为前面的元素:
if (front.data == value) { // If match on first elem
front = current.next;
current = current.next;
}
之后,只需遍历列表并删除匹配的元素。用下面的代码替换有问题的代码应该可以完成以下工作:
while (front.data == value) { // If match on first elem
front = front.next;
if (front == null) {
return;
}
}
从自定义链表中删除元素的最简单、最干净的方法之一是:
import java.util.concurrent.atomic.AtomicInteger;
public class LinkedList<T> {
class Node<T> {
private T data;
private Node<T> next;
Node(T data) {
this.data = data;
this.next = null;
}
}
private Node<T> head;
private Node<T> tail;
private AtomicInteger size = new AtomicInteger();
public void add(T data) {
Node<T> n = new Node<T>(data);
if (head == null) {
head = n;
tail = n;
size.getAndIncrement();
} else {
tail.next = n;
tail = n;
size.getAndIncrement();
}
}
public int size() {
return size.get();
}
public void displayElement() {
while (head.next != null) {
System.out.println(head.data);
head = head.next;
}
System.out.println(head.data);
}
public void remove(int position) throws IndexOutOfBoundsException {
if (position > size.get()) {
throw new IndexOutOfBoundsException("current size is:" + size.get());
}
Node<T> temp = head;
for (int i = 0; i < position; i++) {
temp = temp.next;
}
head = null;
head = temp;
size.getAndDecrement();
}
}