package com.github.mrwengq.tid.list;
public class LinkedList implements List {
private Node head;
private int size =0;
private static class Node {
Object data;
Node next;
public Node(Object o) {
data = o;
next = null;
}
}
public void add(Object o) {
if (size == 0) {
head = new Node(o);
} else {
Node node = new Node(o);
Node lastNode = findNode(size-1);
lastNode.next = node;
}
size++;
}
private Node findNode(int index) {//用于查找节点
Node no = head;
for (; index > 0; index--)
no = no.next;
return no;
}
public void add(int index, Object o) {
if (index < 0 || index > size - 1)
throw new ArrayIndexOutOfBoundsException();
Node node = new Node(o);
Node indexNode = findNode(index);
if (index - 1 < 0) {
node.next = indexNode;
head = node;
size++;
return;
} else {
Node lastNode = findNode(index - 1);
lastNode.next = node;
node.next = indexNode;
size++;
return;
}
}
public Object get(int index) {
if (index < 0 || index > size - 1)
throw new ArrayIndexOutOfBoundsException();
else
return findNode(index).data;
}
public Object remove(int index) {
if (index < 0 || index > size - 1 || size == 0)
throw new ArrayIndexOutOfBoundsException();
Node indexNode = findNode(index);
if (size == 1) {
head = null;
size = 0;
return indexNode.data;
}
Node nextNode = null;
Node lastNode = null;
if (index + 1 <= size - 1) //判断是否有下一位
nextNode = findNode(index + 1);
if (index - 1 >= 0) //判断是否有上一位
lastNode = findNode(index - 1);
if (lastNode == null) {
head = nextNode;
size--;
return indexNode.data;
}else if (nextNode == null) {
lastNode.next = null;
size--;
return indexNode.data;
} else {
lastNode.next = nextNode;
size--;
return indexNode.data;
}
}
public int size() {
return size;
}
public void addFirst(Object o) {
Node node = new Node(o);
if (size == 0) {
head = node;
size++;
return;
} else {
node.next = head;
head = node;
size++;
return;
}
}
public void addLast(Object o) {
Node node = new Node(o);
if (size == 0) {
head = node;
size++;
return;
} else {
Node lastNode = findNode(size-1);
lastNode.next = node;
size++;
return;
}
}
public Object removeFirst() {
if (size == 0) {
return null;
} else {
Node nextNode = head.next;
Object ob = head.data;
head = nextNode;
size--;
return ob;
}
}
public Object removeLast() {
if (size == 0) {
return null;
} else {
Node node = findNode(size-1); //size -1 为最后一位 -2为前一位
if(size-2>=0){
Node lastNode = findNode(size - 2);
lastNode.next = null;
}
size--;
return node.data;
}
}
public Iterator iterator() {
return new Iterator() {
int index = -1;
public boolean hasNext() {
index++;
if(index<size){
Object ob = findNode(index);
return true;
}
return false;
}
public Object next() {
return findNode(index).data;
}
};
}
/**
* 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(size==0){
throw new RuntimeException();
}
int cs = size/2;
int endIndex = size -1;
for(int i = 0;i<cs;i++){
Node node1 = findNode(i);
Node node2 = findNode(endIndex-i);
Node BeforNode1 = null; //node1前一个节点
Node AfterNode2 = null;//node2 后一个节点
if(i != 0 ){
BeforNode1 = findNode(i-1);
AfterNode2 = findNode(endIndex-i).next;
}
Node AfterNode1 = findNode(i).next;
Node BeforNode2 = findNode(endIndex-i-1);
if(BeforNode1!= null && AfterNode2!=null){
BeforNode1.next = node2;
node1.next = AfterNode2;
}
BeforNode2.next = node1;
node2.next = AfterNode1;
if(i==0){
head = node2;
}
}
}
/**
* 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){
throw new RuntimeException();
}
int len = size/2;
Node node = findNode(len-1);//len-1为删除链表的最后一位下标
for( int j = len-2; j>=0 ; j--){
Node temp = findNode(j);
temp.next = null;
}
head = node.next;
node.next = null;
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 i, int length){
if(0 == size||i > size-1){
throw new RuntimeException();
}
Node beforNode = findNode(i-1); //i的前一个元素
Node afterNode = findNode(i+length);//被删除最大下标节点的下一个节点
for( int j = i+length-1; j<i ; j--){
Node temp = findNode(j);
temp.next = null;
}
beforNode.next = afterNode;
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){
int temp = list.size()-1;
if(temp>size){
throw new RuntimeException();
}
int[] b = new int[list.size()];
for(int i = 0;i<list.size;i++){
temp = (int)list.get(i);
Node no = head;
while(temp>0){
for (; temp > 0; temp--)
no = no.next;
}
b[i] = (int)no.data;
}
return b;
}
/**
* 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){
if(list==null){
throw new RuntimeException();
}
for(int i = 0;i<list.size();i++){
for(int j = 0;j<this.size();j++){
if((int)this.get(j)==(int)list.get(i)){
System.out.println(this.get(j)+" " +j);
this.remove(j);
break;
}
}
}
}
/**
* 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(){
int len = size -1;
for(int i = 0;i<len;i++){
int next = i+1;
while((int)this.get(i)==(int)this.get(next)){
this.remove(next);
len = this.size()-1;
}
}
}
/**
* 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){
int lmin = 0;
int lmax = size-1;
int lmid = (int)size-1/2;
int rmin = 0;//在需要删除的集合中的最小值
int rmax = 0;//在需要删除的集合中的最大值
while(lmin<lmax){
lmid = (lmax+lmin)/2;
if((int)this.get(lmid) > min){
lmax = lmid-1;
if((int)this.get(lmax)<min){
rmin = lmid;
break;
}
}else if((int)this.get(lmid) <= min){
lmin = lmid+1;
if((int)this.get(lmin)>min){
rmin = lmin;
break;
}
}
}
while(lmin<lmax){
lmid = (lmax+lmin)/2;
if((int)this.get(lmid)<max){
lmin = lmid +1;
if((int)this.get(lmin)>max){
rmax = lmid;
}
}else if((int)this.get(lmid)>=max){
lmax = lmid -1;
if((int)this.get(lmax)<max){
rmax = lmax;
}
}
}
Node beforNode = findNode(rmin-1);
Node afterNode = findNode(rmax+1);
for(int i = rmax;i>=rmin;i--){
Node removeNode = findNode(i);
removeNode.next = null;
size--;
}
beforNode.next = afterNode;
}
/**
* 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){
int len = size;
int llen = list.size();
int i = 0;
int j = 0;
LinkedList ll= new LinkedList();
while(true){
if(i == len &&j == llen ){
break;
}
if(i>len-1){
ll.add(list.get(j++));
continue;
}
if(j>llen-1){
ll.add(this.get(i++));
continue;
}
if((int)get(i)<(int)list.get(j)){
ll.add(this.get(i++));
}else if((int)get(i)>(int)list.get(j)){
ll.add(list.get(j++));
}else{
ll.add(list.get(j++));
i++;
}
}
return ll;
}
}