package com.coding.basic;
import javax.xml.crypto.Data;
public class LinkedList implements List {
private Node head;
private int length;
//构造函数
public LinkedList(){
clear();
}
public final void clear(){
head = null;
length = 0;
}
public void add(Object o){
Node newNode = new Node(o);
if(length == 0)
{
head = newNode;
}
else{
Node lastNode = getNodeAt(length);
lastNode.next = newNode;
}
length++;
}
public void add(int index , Object o){
Node newNode = new Node(o);
Node nodeBefor = getNodeAt(index-1);
Node nodeAfter = nodeBefor.next;
newNode.next = nodeAfter;
nodeBefor.next = newNode;
length++;
}
public Object get(int index){
if((1<=index)&&(index<=length))
{
Node currentNode = head;
for(int i= 0;i<index;i++)
{
currentNode = currentNode.next;
}
return currentNode.data;
}
else
return null;
}
public Object remove(int index){
Node nodeBefor = getNodeAt(index-1);
Node nodeToRemove = nodeBefor.next;
Node nodeAfter = nodeToRemove.next;
nodeBefor.next = nodeAfter;
length--;
return nodeToRemove.data;
}
public int size(){
return length;
}
public void addFirst(Object o){
Node newNode = new Node(o);
newNode.next = head;
head = newNode;
length++;
}
public void addLast(Object o){
Node newNode = new Node(o);
Node lastNode = getNodeAt(length);
lastNode.next = newNode;
length++;
}
public Object removeFirst(){
head = head.next;
length--;
return head.data;
}
public Object removeLast(){
Node nodeLastBefore = getNodeAt(length-1);
Object result = nodeLastBefore.next.data;
nodeLastBefore.next =null;
length--;
return result;
}
public Iterator iterator(){
return null;
}
private Node getNodeAt(int givenPosition){
if((1<=givenPosition)&&(givenPosition <=length)){
Node currentNode = head;
for (int counter = 1;counter<givenPosition;counter++)
currentNode = currentNode.next;
return currentNode;
}else
return null;
}
private static class Node{
private Object data;
private Node next;
private Node(Object dataPortion){
data = dataPortion;
next = null;
}
private Node(Object dataPortion,Node nextNode)
{
data = dataPortion;
next = nextNode;
}
public Node() {
}
}
/**
* 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(){
Node lastNode = getNodeAt(length);
head = lastNode;
while(length>0){
Node currentNode = getNodeAt(--length);
add(currentNode);
}
}
/**
* 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(){
int num = length/2;
while(num>0){
remove(num);
num--;
}
}
/**
* 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){
while (length>0){
remove(i+length);
length--;
}
}
/**
* 假定当前链表和list均包含已升序排列的整数
* 从当前链表中取出那些list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){
int[] arr = new int[list.size()];
for(int i =0;i<list.size();i++){
arr[i] =(int) this.get((int)list.get(i));
}
//如何操作当前链表？
return arr;
}
/**
* 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list中出现的元素
* @param list
*/
public void subtract(LinkedList list){
for (int i = 0; i < list.size(); i++) {
this.remove((int)list.get(i));
}
}
/**
* 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(){
for(int i =0;i<length;i++){
Node currentNode = getNodeAt(i);
for(int j =i;j<length;j++){
Node compareNode = getNodeAt(j);
if(currentNode.data == compareNode.data)
remove(i);
}
}
}
/**
* 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){
for(int i = 0;i<length;i++){
int data = (int)getNodeAt(i).data;
if (min<data&&max>data)
remove(i);
}
}
/**
* 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){
if(list==null){
return null;
}
int i1 = 0;
int i2 = 0;
LinkedList result = new LinkedList();
Node currentListNode = list.head;
Node currentThisNode = this.head;
for(i1 =0;i1<list.size();i1++){
for(i2 = 0;i2<this.size();i2++){
if(list.getNodeAt(i1)==this.getNodeAt(i2))
result.add(list.getNodeAt(i1));
break;
}
}
return result;
}
}