package week1.collections;
public class LinkedList implements List {
private int size = 0;
private Node head;
private Node last;
public boolean add(Object o){
Node newNode = new Node(o);
if(head == null){
last = newNode;
head = newNode;
}else{
Node oldLast = last;
last = newNode;
oldLast.next = last;
}
size++;
return true;
}
public void add(int index , Object o){
outOfIndex(index);
if(index == 0){
Node oldHead = head;
head = new Node(o);
head.next = oldHead;
}else{
Node h = getNode(index-1);
Node newNode = new Node(o);
newNode.next = h.next;
h.next = newNode;
}
size++;
}
public Object get(int index){
Node h = getNode(index);
return h.data;
}
private Node getNode(int index) {
outOfIndex(index);
Node h = head;
for(int i=0;i<index;i++){
h = h.next;
}
return h;
}
private void outOfIndex(int index) {
if(index >= size || index < 0){
throw new IndexOutOfBoundsException("Index"+index+"越界");
}
}
public Object remove(int index){
outOfIndex(index);
Object data;
if(index==0){
Node oldHead = head;
head = head.next;
data = oldHead.data;
oldHead = null;
}else{
Node preNode = getNode(index-1);
if(preNode.next==last){
Node oldLast = last;
last = preNode;
data = oldLast.data;
oldLast = null;
}else{
Node removeNode = preNode.next;
preNode.next = preNode.next.next;
data = removeNode.data;
removeNode = null;
}
}
size--;
return data;
}
public int size(){
return size;
}
public void addFirst(Object o){
add(0,o);
}
public void addLast(Object o){
if(last==null){
add(o);
}else{
Node oldLast = last;
last = new Node(o);
oldLast.next = last;
}
size++;
}
public Object removeFirst(){
return remove(0);
}
public Object removeLast(){
return remove(size-1);
}
public Iterator iterator(){
return new LinkedListIterator();
}
private class LinkedListIterator implements Iterator{
int pos = 0;
@Override
public boolean hasNext() {
return pos<size();
}
@Override
public Object next() {
return get(pos++);
}
}
private static class Node{
Object data;
Node next;
public Node(){
}
public Node(Object o){
this.data = o;
this.next = null;
}
}
/**
* 把该链表逆置
* 例如链表为 3->7->10 , 逆置后变为 10->7->3
*/
public void reverse(){
}
/**
* 删除一个单链表的前半部分
* 例如:list = 2->5->7->8 , 删除以后的值为 7->8
* 如果list = 2->5->7->8->10 ,删除以后的值为7,8,10
*/
public void removeFirstHalf(){
}
/**
* 从第i个元素开始, 删除length 个元素 , 注意i从0开始
* @param i
* @param length
*/
public void remove(int i, int length){
}
/**
* 假定当前链表和listB均包含已升序排列的整数
* 从当前链表中取出那些listB所指定的元素
* 例如当前链表 = 11->101->201->301->401->501->601->701
* listB = 1->3->4->6
* 返回的结果应该是[101,301,401,601]
* @param list
*/
public int[] getElements(LinkedList list){
return null;
}
/**
* 已知链表中的元素以值递增有序排列,并以单链表作存储结构。
* 从当前链表中中删除在listB中出现的元素
* @param list
*/
public void subtract(LinkedList list){
}
/**
* 已知当前链表中的元素以值递增有序排列,并以单链表作存储结构。
* 删除表中所有值相同的多余元素(使得操作后的线性表中所有元素的值均不相同)
*/
public void removeDuplicateValues(){
}
/**
* 已知链表中的元素以值递增有序排列,并以单链表作存储结构。
* 试写一高效的算法,删除表中所有值大于min且小于max的元素(若表中存在这样的元素)
* @param min
* @param max
*/
public void removeRange(int min, int max){
}
/**
* 假设当前链表和参数list指定的链表均以元素依值递增有序排列(同一表中的元素值各不相同)
* 现要求生成新链表C,其元素为当前链表和list中元素的交集,且表C中的元素有依值递增有序排列
* @param list
*/
public LinkedList intersection( LinkedList list){
return null;
}
}