package com.coding.basic.third;
import java.util.Iterator;
public class LinkedList implements List {
private Node head;
private int size;
public void add(Object o){
add(size,o);
}
public void add(int index , Object o){
if(size<0 || size < index){
new ArrayIndexOutOfBoundsException();
}
Node node = new Node();
node.data = o;
if(head==null){
head = node;
}else{
Node next = head.next;
if(next == null){
head.next = node;
}else{
while(next.next!=null){
next = next.next;
}
next.next = node;
}
}
size++;
}
public Object get(int index){
if(size<0 || size < index){
new ArrayIndexOutOfBoundsException();
}
if(index == 0){
return head.data;
}
int temp = 0;
Node next = head;
while(index > temp){
temp++;
next = next.next;
}
return next.data;
}
public Object remove(int index){
if(index<0 || size < index){
new ArrayIndexOutOfBoundsException();
}
Node res = head;
if(index == 0){
head = head.next;
size--;
return res.data;
}
int temp = 1;
Node next = head;
while(index > temp){
temp++;
next = next.next;
}
res = next.next;
next.next = next.next.next;
size--;
return res.data;
}
public int size(){
return size;
}
public void addFirst(Object o){
add(0,o);
}
public void addLast(Object o){
add(o);
}
public Object removeFirst(){
Node result = head;
head = head.next;
return result.data;
}
public Object removeLast(){
return remove(size-1);
}
public Iterator iterator(){
return null;
}
private static class Node{
Object data;
Node next;
}
/**
* 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(){
Object[] obj = new Object[size];
int tem = 0;
Node next = head;
while(size>tem){
obj[tem] = next.data;
next = next.next;
tem++;
}
Node nextF = head;
for(int k=obj.length-1;k>=0;k--){
nextF.data = obj[k];
nextF = nextF.next;
}
}
/**
* 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(){
if(head == null || head.next ==null){
return;
}
int len = size/2;
int temp = 0;
Node next = head;
while(len<temp){
temp++;
next = next.next;
}
head = next;
head.next = next.next;
size = size-len;
}
/**
* 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 index, int length){
if(index<0 || size < index){
new ArrayIndexOutOfBoundsException();
}
int temp = 1;
Node next = head;
while(index > temp){
temp++;
next = next.next;
}
Node nextF = next;
int len = length;
while(length>0){
length--;
next = next.next;
}
nextF.next = next.next;
size = size-len;
}
/**
* 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){
int[] ret = new int[list.size()];
for(int i=0;i<ret.length;i++){
ret[i] = (int) get((int)list.get(i));
}
return ret;
}
/**
* 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(){
Node next = head;
while(next!= null && next.next!=null){
if(next.data == next.next.data){
next.next = next.next.next;
size--;
}
}
}
/**
* 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){
Node next = head;
if (next == null || (int)next.data >= max) {
return;
}
Node start = head;
while(next!=null && (int)next.data < min){
start = next;
next = next.next;
}
Node end = head;
while(next!=null && (int)next.data < max){
end = next;
next = next.next;
}
if(start ==null){
head = end;
}else{
start.next = end.next;
}
}
/**
* 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){
LinkedList linkedList = new LinkedList();
int i = 0;
Node next = head;
Node nextF = list.head;
while (next != null || nextF != null ){
if (nextF != null && next != null) {
if ((int)next.data < (int)nextF.data ) {
linkedList.add(next.data);
next = next.next;
} else if(next == null ||(int)nextF.data < (int)next.data) {
linkedList.add(nextF.data);
nextF = nextF.next;
} else {
linkedList.add(nextF.data);
next = next.next;
nextF = nextF.next;
}
}else {
if (nextF == null) {
linkedList.add(next.data);
next = next.next;
} else {
linkedList.add(nextF.data);
nextF = nextF.next;
}
}
}
return linkedList;
}
}