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Merge pull request #2 from cjl1407/master
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第一次作业
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dsfan authored Mar 10, 2017
2 parents 1c07719 + 1203ca2 commit bb70879
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42 changes: 42 additions & 0 deletions group22/1158477486/src/TestCollection/ArrayList.java
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package TestCollection;

public class ArrayList implements List {



private Object[] elementData = new Object[100];
private int size ;
public void add(Object o){
if(size==elementData.length){
Object[]newData=new Object[size*2+1];//½«Êý×éÀ©ÈÝ
System.arraycopy(elementData, 0, newData, 0, elementData.length);
elementData=newData;
}
elementData[size++]=o;

}
public void add(int index, Object o){
System.arraycopy(elementData, index, elementData,index+1 , size-index);
elementData[index]=o;
size++;
}

public Object get(int index){
return elementData[index];
}

public Object remove(int index){
System.arraycopy(elementData, index+1, elementData, index, size-index-1);
elementData[--size]=null;
return null;
}

public int size(){
return size;
}

public Iterator iterator(){
return null;
}

}
184 changes: 184 additions & 0 deletions group22/1158477486/src/TestCollection/LinkedList.java
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package TestCollection;



public class LinkedList implements List {

private Node head=new Node();;
private int size;
public Node node(int a){//创建得到a位置节点的方法
Node teamp=null;

if(head!=null){
if(a<=size ){
teamp=head;
for(int i=0;i<a;i++){
teamp=teamp.next;//得到第a个位置的节点
}
}
}return teamp;
}
public void add(Object o){

if(head .next== null){
Node firstNode=new Node();//形成第一个节点

head.data=null;
head.next=firstNode;//将头结点指向第一个节点
firstNode.data=o;
firstNode.next=null;
}else{
Node firstNode= head.next;//形成第一个节点
Node newNode=new Node();
newNode.data=o;
newNode.next=null;
firstNode.next=newNode;//把形成的新节点放在最后一个节点后面
firstNode=newNode;//使新形成的节点变成最后一个节点

}
size++;
}
public void add(int index , Object o){

Node teamp=node(index);//得到第index位置上的节点
Node teamp1=node(index-1);//得到teamp前一个位置的节点
Node newNode=new Node();//创建新节点
newNode.data=o;
newNode.next=teamp;
teamp1.next=newNode;
size++;
}
public Object get(int index){
Node teamp=node(index);//得到第index位置上的节点

Object o= teamp.data;
return o ;
}
public Object remove(int index){
Node teamp=node(index);//得到第index位置上的节点
Node teamp1=node(index-1);//得到第index-1位置上的节点
Node teamp2=node(index+1);//得到第index+1位置上的节点
teamp=null;
teamp1.next=teamp2;//将teap1下一个位置指向teamp2
size--;
return null;
}

public int size(){

return size;
}

public void addFirst(Object o){
Node firstNode=new Node();
firstNode.data=o;

Node teamp=node(0);//得到原来的第一个节点
firstNode.next=teamp;//在第一个位子插入节点
head.next=firstNode;
size++;
}
public void addLast(Object o){
Node lastNode=new Node();
lastNode.data=o;

Node teamp=node(size );//得到原来的最后节点
teamp.next=lastNode;//在最后位子插入节点
lastNode.next=null;
size++;

}
public Object removeFirst(){
return null;
}
public Object removeLast(){
return null;
}
public Iterator iterator(){
return null;
}


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(){

}

/**
* 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(){

}

/**
* 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){

}
/**
* 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 static int[] getElements(LinkedList list){
return null;
}

/**
* 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(){

}

/**
* 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){

}

/**
* 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){
return null;
}
}


9 changes: 9 additions & 0 deletions group22/1158477486/src/TestCollection/List.java
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package TestCollection;

public interface List {
public void add(Object o);
public void add(int index, Object o);
public Object get(int index);
public Object remove(int index);
public int size();
}
38 changes: 38 additions & 0 deletions group22/1158477486/src/TestCollection/Queue.java
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package TestCollection;

;

public class Queue {
private LinkedList list=new LinkedList();

public void enQueue(Object o){
list.addLast ( o) ;//ÎÊÌâÓÃadd(index, o)³ö´í
}

public Object deQueue(){
Object o=list.get(1);
list.remove(1);
return o;
}

public boolean isEmpty(){
if(list.size()!=0){
return false;}else
return true;
}

public int size(){
return list.size();
}
public static void main(String[] args) {
Queue q=new Queue();
q.enQueue("1");
q.enQueue("2");
q.enQueue("3");
System.out.println(q.size());
Object o=q.deQueue();
System.out.println(o);
System.out.println(q.size());

}
}
36 changes: 36 additions & 0 deletions group22/1158477486/src/TestCollection/Stack.java
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package TestCollection;

public class Stack {
private ArrayList elementData = new ArrayList();

public void push(Object o){
elementData.add(o);
}

public Object pop(){
return elementData.get(elementData.size()-1) ;
}

public Object peek(){
elementData.remove(elementData.size()-1);
return elementData.get(elementData.size()-1);
}
public boolean isEmpty(){
if(elementData.size()!=0){
return false;}
return true;
}
public int size(){
return elementData.size();
}
public static void main(String[] args) {
Stack s=new Stack();
s.push("111");
s.push("211");
s.push("311");
System.out.println(s.size());//3
System.out.println(s.pop());
System.out.println(s.size());
System.out.println(s.peek());
System.out.println(s.size());
}}

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