package com.coding.basic;
import java.util.Stack;
public class LinkedList implements List {
private int size;
private Node head;
public LinkedList(){
size = 0;
head = null;
}
public void add(Object o){
Node nd = new Node(o);
if(head == null){
head = nd;
}else{
Node p = head;
while(p.next != null){
p = p.next;
}
p.next = nd;
}
size ++;
}
//3 > 2 > 1 > 5 > 4 > 5改变之前
//0 1 2 3 4 5index
//3 > 2 > 1 > x > 5 > 4 > 5插入之后
public void add(int index , Object o){
if(head != null){
int k = 0;
Node p = head;
while(k < index - 1 && p.next != null){
k++;
p = p.next;//当前p为要插入位置的前一个节点
}
if(p != null){
Node nd = new Node(o);
nd.next = p.next;
p.next = nd;
}
size++;
}
}
public Object get(int index){
if(index < 0 || index >= size){
throw new IndexOutOfBoundsException();
}
Node p = head;
int k = 0;
while(k < index && p.next !=null){
k++;
p = p.next;
}
return p.data;
}
//3 > 2 > 1 > 5 > 4 > 5改变之前
//0 1 2 3 4 5index
//3 > 2 > 1 > 4 > 5插入之后
public Object remove(int index){
if(index < 0 || index >= size){
throw new IndexOutOfBoundsException();
}
if(head == null){
return null;
}
if(index == 0){
head = head.next;
size--;
return head.data;
}else{
if(head != null){
int k = 0;
Node p = head;
while(k < index - 1 && p != null){
k++;
p = p.next;
}
Node pn = p.next;
if(pn != null){
p.next = pn.next;
size--;
return pn.data;
}
}
}
return null;
}
public int size(){
return size;
}
public void addFirst(Object o){
if(head != null){
Node nd = new Node(o);
Node first = head;
head = nd;
first = nd.next;
}
}
public void addLast(Object o){
if(head != null){
int k = 0;
Node p = head;
while(p.next != null && k < size - 1){
p = p.next;
k++;
}
Node newNode = new Node(o);
p.next = newNode;
}
}
public Object removeFirst(){
Node node = head;
if(head != null){
head = head.next;
}
return node.data;
}
public Object removeLast(){
Node p = head;
int k = 0;
while(p.next != null && k < size - 2){
k++;
p = p.next;
}
p.next = null;
return p.next;
}
public Iterator iterator(){
return null;
}
private static class Node{
Object data;
Node next;
private Node(Object o){
this.data = o;
this.next = null;
}
}
/**
* 把该链表逆置
* head head
* 例如链表为 3->7->10 , 逆置后变为 10->7->3
*/
public void reverse(){
/**
* 长度超过1的单链表需要逆转
*/
if(head == null || head.next == null){
return;
}
Stack<Node> st = new Stack<Node>();
Node currentNode = head;
while(currentNode != null){
st.push(currentNode);
Node nextNode = currentNode.next;
currentNode.next = null;//断开连接
currentNode = nextNode;
}
head = (Node) st.pop();
currentNode = head;
while(!st.isEmpty()){
Node nextNode = (Node) st.pop();
currentNode.next = nextNode;
currentNode = nextNode;
}
}
/**
* 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 = size / 2;
for(int i = 0; i < num; 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){
throw new IndexOutOfBoundsException();
}
int len = size - i >= length ? length : size - i;
int k = 0;
while(k < len){
remove(i);
k++;
}
}
/**
* 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[] newList = new int[list.size()];
for(int i = 0;i < list.size(); i++){
newList[i] = Integer.parseInt(this.get(Integer.parseInt(list.get(i).toString())).toString());
}
return newList;
}
/**
* 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){
for(int j = 0;j < list.size();j++){
this.remove(list.get(j));
}
}
/**
* 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){
throw new RuntimeException("LinkedList is null");
}
Node currentNode = head;
Node preNode = head;
while(currentNode.next != null){
currentNode = currentNode.next;
Object data = preNode.data;
while(currentNode.data == data){
if(currentNode.next == null){
preNode.next = null;
break;
}
preNode.next = currentNode.next;
size--;
currentNode = currentNode.next;
if(currentNode == null){
break;
}
}
preNode = preNode.next;
}
}
/**
* 传入删除数据节点
*/
public void remove(Object obj){
if(head == null){
throw new RuntimeException("linkedlist is nuull");
}
if(head.data.equals(obj)){
head = head.next;
size--;
}else{
Node pre = head;
Node currentNode = head.next;
while(currentNode != null){
if(currentNode.data.equals(obj)){
pre.next = currentNode.next;
size--;
}
pre = pre.next;
currentNode = currentNode.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){
Node node = head;
int start = -1;
int end = -1;
int i = 0;
while(node != null){
if((Integer)node.data <= min){
start = i;
}
if((Integer)node.data >= max){
end = i;
break;
}
node = node.next;
i++;
}
if(start == -1){
start = 0;
}
if(end == -1){
end = size;
}
this.remove(start+1, end-start-1);
}
/**
* 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;
}
LinkedList newList = new LinkedList();
int fi = 0;
int se = 0;
while(fi < this.size && se < list.size()){
int val1 = (Integer) this.get(fi);
int val2 = (Integer) list.get(se);
if(val1 == val2){
newList.add(val1);
fi++;
se++;
}else if(val1 < val2){
fi++;
}else{
se++;
}
}
return newList;
}
public static void main(String[] args) {
LinkedList linkedList = new LinkedList();
linkedList.add(11);
linkedList.add(22);
linkedList.add(33);
linkedList.add(44);
linkedList.add(55);
linkedList.reverse();
for(int i = 0; i < linkedList.size; i++){
System.out.println(linkedList.get(i));
}
}
}