package com.oneflyingleaf.util;
public class LinkedList implements List {
private Node head;
//虚拟tail节点,方便add最后节点,减少成,tail为链表的最后一个元素
private Node tail;
private int size ;
public void add(Object o){
size ++ ;
if(head == null){
head = new Node();
head.data = o;
tail = head;
return ;
}
Node node = tail;
tail = new Node();
node.next = tail;
tail.data = o;
}
public void add(int index , Object o){
checkBound(index,true);
if(index == 0){
addFirst(o);
return ;
}
if(index == (size - 1)){
add(o);
return ;
}
size ++;
Node node = head;
while(index-- > 0){
node = node.next;
}
Node temp = new Node();
temp.data = o;
temp.next = node.next;
node.next = temp;
}
public Object get(int index){
checkBound(index,false);
Node node = head;
while(index > 0){
index -- ;
node = node.next;
}
return node.data;
}
public Object remove(int index){
checkBound(index,false);
if(index == 0){
removeFirst();
}
if(index == (size-1)){
removeLast();
}
size --;
Node node = head;
while(--index > 0){
node = node.next;
}
Object o = node.next.data;
node.next = node.next.next;
return o;
}
public int size(){
return size ;
}
public void addFirst(Object o){
size ++ ;
Node node = new Node();
node.data = o;
node.next = head;
head = node;
}
public void addLast(Object o){
add(o);
}
public Object removeFirst(){
checkBound(0,true);
if(size == 0){
//处理tail节点,防止占用引用资源回收不了
tail.data = null;
}
size -- ;
Object o = head.data;
head = head.next;
return o;
}
public Object removeLast(){
if(size == 0){
//直接移除首节点,则无需处理尾节点
removeFirst();
}
size -- ;
Object o = tail.data;
tail = null;
Node temp = head ;
while(temp != null){
temp = temp.next;
}
tail = temp;
return o;
}
public Iterator iterator(){
return new LinkedListIterator();
}
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(){
Node node = head.next;
Node pre = head;
Node next;
while(node.next != null){
next = node.next;
node.next = pre;
pre = node;
node = next;
}
//处理尾元素
node.next = pre;
next = tail;
tail = head;
head = next;
tail.next = null;
}
/**
* 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 size = this.size / 2;
for(int i = 0;i < size ; 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){
for(int j = i;length > 0; length -- ,j++ ){
remove(j);
}
}
/**
* 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 Object[] getElements(LinkedList list){
Object [] ret = new Object[list.size];
for(int i = 0 ;i < list.size;i++){
ret[i] = 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list中出现的元素
* @param list
*/
public void subtract(LinkedList list){
int size = this.size > list.size()?this.size : list.size();
int temp = 0;
for(int i = 0;i < size ;i++){
if(temp == size || i == list.size){
return ;
}
if((int)list.get(i) > (int)get(temp)){
temp ++;
}else if((int)list.get(i) < (int)get(temp)){
continue;
}else{
remove(temp);
}
}
}
/**
* 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(size == 0){
return ;
}
Object o = get(0);
for(int i = 0;i < (size - 1) ;i++){
if(o.equals(get(i + 1))){
remove(i + 1);
i--;
}
o = 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(size == 0 || max < (int)get(0) || min > (int)get(size - 1)){
return ;
}
int minIndex = getIndex(min,0,size);
int maxIndex = getIndex(max,0,size);
remove(minIndex,maxIndex - minIndex );
}
/**
* 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 ret = new LinkedList();
int temp = 0;
for(int i = 0;i < size ;i++){
if(temp == size || i == list.size){
return ret ;
}
if((int)list.get(i) > (int)get(temp)){
temp ++;
}else if((int)list.get(i) < (int)get(temp)){
continue;
}else{
ret.add(list.get(i));
}
}
return ret ;
}
private class LinkedListIterator implements Iterator{
Node node = null;
@Override
public boolean hasNext() {
if(node == null){
return head != null;
}
return node.next != null;
}
@Override
public Object next() {
if(node == null){
node = head;
}else{
node = node.next;
}
return node.data;
}
}
/**
* 数组越界提示
* @param index 数组下标
* @param contailLast
*/
private void checkBound(int index , boolean containLast){
if(containLast && index == (size + 1) ){
return ;
}
if(index < 0 || index >= size){
throw new IndexOutOfBoundsException("idnex:" + index +", size:" + size);
}
}
/**
* 获取不大于val的下标
* @return
*/
private int getIndex(int val,int min ,int max){
int mid = (0 + size)/2 ;
if(max <= min){
return min;
}
if(val > (int)get(mid)){
mid = (mid + 1 + max) /2 ;
return getIndex(val,mid,max);
}else if(val < (int)get(mid)){
mid = (mid + min -1) /2;
return getIndex(val,min,mid);
}else{
return mid;
}
}
}