package com.LinkedList;
import java.util.Iterator;
import java.util.LinkedList;
public class LinkedListUtil<T extends Comparable> {
private Node<T> head;
private Node<T> current;
private int size = 0;
public void add(T o){
if(head == null){
head = new Node<T>(o);
current = head;
}else{
Node<T> tmp = new Node<T>(o);
current.next = tmp;
current = tmp;
}
size++;
}
public int[] toIntArray(){
if(size == 0) return null;
int[] intArray = new int[this.size()];
Node<T> tmp = head;
for(int i=0; i<size; i++){
intArray[i] = (int)tmp.data;
tmp = tmp.next;
}
return intArray;
}
public void add(int index , Object o){
}
public Node<T> get(int index){
checkBounds(index);
if (index == 0) {
return head;
} else {
Node<T> pos = findIndexPosition(index);
return pos;
}
}
public Object remove(int index){
return null;
}
public void remove(Node<T> n){
Node<T> former = head;
while(former.next != n) former = former.next;
former.next = n.next;
size--;
}
public int size(){
return size;
}
public void addFirst(Object o){
}
public void addLast(Object o){
}
public Node<T> removeFirst(){
if(head == null) return null;
Node<T> tmp = head;
head = head.next;
this.size--;
return tmp;
}
public Object removeLast(){
return null;
}
public Iterator iterator(){
return null;
}
public void checkBounds(int index) {
if (index < 0 || index > size - 1) {
// System.out.println("From MyLinkedList: Index out of bounds");
throw new IndexOutOfBoundsException(OutOfBoundsMsg(index));
}
}
public String OutOfBoundsMsg(int index) {
return "Index: " + index + ", Size: " + size;
}
public Node<T> findIndexPosition(int index) {
Node<T> pos = head;
for (int i = 0; i < index; i++) {
pos = pos.next;
}
return pos;
}
private static class Node<T>{
T data;
Node<T> next;
public Node(T 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(){
if(head == null) return;
Node<T> newHead = head;
head = head.next;
newHead.next = null;
while(head != null){
Node<T> tmp = head;
head = head.next;
tmp.next = newHead;
newHead = tmp;
}
head = newHead;
}
/**
* 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 midLen = this.size()/2;
for(int i=0; i<midLen; i++){
removeFirst();
}
}
/**
* 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){
if(i < 0 || i >= size) return;
if(length < 0 ||length >= size) return;
if(i == 0){
for(int j=0; j<length; j++){
removeFirst();
}
}else{
Node<T> startNode = head;
for(int j=0; j<i-1; j++){
startNode = startNode.next;
}
int cnt = 0;
Node<T> pointer = startNode;
while(cnt<length && pointer != null){
pointer = pointer.next;
cnt++;
}
startNode.next = pointer.next;
this.size = size - 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){
Iterator<Integer> it = list.iterator();
int[] restArray = new int[list.size()];
int cnt = 0;
while(it.hasNext()){
int index = it.next();
restArray[cnt++] = (int)get(index).data;
}
return restArray;
}
/**
* 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){
}
/**
* 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(){
if(head == null) return;
Node<T> cur = head.next;
Node<T> former = head;
while(cur != null){
if(former.data.equals(cur.data)){
remove(cur);
cur = cur.next;
}else{
former = former.next;
cur = cur.next;
}
}
}
/**
* 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){
if((int)head.data > min){
head = findMaxLocation(head, max);
}else{
Node<T> former = head;
while((int)former.data < min && former != null){
former = former.next;
}
Node<T> behind = findMaxLocation(former.next, max);
former.next = behind;
}
}
public Node<T> findMaxLocation(Node<T> n, int max){
Node<T> tmp = n;
while((int)tmp.data < max && tmp != null){
tmp = tmp.next;
size--;
}
return tmp;
}
/**
* 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(head == null) return list;
return null;
}
}