package com.bruce.homework0312.linkedlist;
import com.bruce.homework0226.LinkedListV01;
import java.util.Objects;
public class LinkedListV02<T> {
private int size = 0;
private Node<T> head;
private Node<T> last;
public LinkedListV02() {
//保证了初始化一个对象的时候,头节点不为空
this.head = new Node<T>(null);
}
public boolean add (T element) {
//双向链表,双向都需要维护
if(last == null){
last = new Node<>(element);
head.next = last;
last.pre = head;
}else{
Node<T> oldLast = last;
last = new Node<>(element);
last.pre = oldLast;
oldLast.next = last;
}
size++;
return true;
}
public boolean add(int index, T element) {
Node<T> node = getNode(index);
Node<T> newNode = new Node<T>(element);
Node<T> pre = node.pre;
pre.next = newNode;
newNode.pre = pre;
newNode.next = node;
size++;
return true;
}
public boolean remove(T element){
Node<T> node = head;
//下一个节点不为null
while(node.next != null){
node = node.next;
if(Objects.equals(node.element, element)){
if(node.next != null){
node.next.pre = node.pre;
}
node.pre.next = node.next;
size--;
return true;
}
}
//下一个节点为null,说明是尾节点
if(node != head){
last = node;
}
//head.next=null,说明是一个空的链表,即仅有一个空head节点
return false;
}
public T remove(int index){
Node<T> node = getNode(index);
Node<T> pre = node.pre;
Node<T> next = node.next;
pre.next = next;
next.pre = pre;
size--;
return node.element;
}
public void clear(){
for(Node<T> x = head; x != null; ){
Node<T> next = x.next;
x.pre = null;
x.next = null;
x.element = null;
}
head = last = null;
size = 0;
}
public int size(){
return size;
}
public boolean isEmpty(){
return size == 0;
}
public boolean contains(Object o){
for(int i = 0; i < size; i++){
if(Objects.equals(getNode(i).element, o)){
return true;
}
}
return false;
}
public Node<T> getNode(int index){
if(index < 0 || index >size){
return null;
}
Node<T> node = head;
for(int i = 0; i < index; i++){
node = node.next;
}
return node;
}
public T get(int index) {
return getNode(index).element;
}
public int indexOf(T element){
Node<T> node = head;
int index = 0;
while(node.next != null){
node = node.next;
if(Objects.equals(node.element, element)){
return index;
}
index++;
}
return -1;
}
private static class Node<T> {
T element;
Node<T> pre;
Node<T> next;
Node(T element) {
this.element = element;
}
}
/**
* 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(){
T t;
for(int i = 0; i < size; i++) {
t = getNode(i).element;
getNode(i).element = getNode(size-1-i).element;
getNode(size-1-i).element = t;
}
}
/**
* 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(size < 2) {
return;
}
int half = size >> 1;
for(int i = 0; i < half; i++) {
remove(i);
}
}
/**
* 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+length)<0 || (i+length)>size) {
return;//抛出异常
}
for(int n = i - 1; n <= length; n++) {
remove(n);
}
}
/**
* 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(LinkedListV02<Integer> list){
if(list == null) {
return null;
}
int[] result = new int[list.size()];
for(int i = 0; i < list.size(); i++) {
if(list.get(i) < size) {
result[i] = (Integer) this.get(i);
}
}
return result;
}
/**
* 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(LinkedListV02 list){
if(list == null || list.size() == 0) {
return;
}
for(int i = 0; i < list.size(); i++) {
if(this.contains(list.get(i))) {
this.remove((T) 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(){
LinkedListV02<T> newList = new LinkedListV02<>();
for(int i = 0; i < size; i++) {
if(!newList.contains(this.get(i))) {
newList.add(this.get(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){
if(min > max) {
return;
}
if((Integer)head.element > min && (Integer)last.element < max) {
clear();
} else if ((Integer)head.element > min && (Integer)last.element > max) {
Node<T> temp1 = last;
Node<T> temp2 = head;
while(temp1.pre != null) {
temp1 = temp1.pre;
if(Objects.equals(temp1.element, max)) {
last = temp1;
} else {
temp1.pre = null;
temp1.element = null;
}
temp1.next = null;
}
while(temp2.next != null) {
temp2 = temp2.next;
if(Objects.equals(temp2.element, min)) {
head = temp2;
} else {
temp2.next = null;
temp2.element = null;
}
temp2.pre = null;
}
}
}
/**
* 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 LinkedListV02 intersection(LinkedListV02 list){
if(list == null || list.size() == 0) {
return null;
}
LinkedListV02 newList = new LinkedListV02();
for(int i = 0; i < list.size(); i ++) {
if(this.contains(list.get(i))) {
newList.add(list.get(i));
}
}
return newList;
}
}