csinglelink

package csinglelink;

class Node {

int iData;
Node Child;

public Node() {
}

public Node(int i) {
iData = i;
Child = null;
}
}

class SingleLink {

Node head = null;

public SingleLink() {
}
//展示链表
public void display() {
Node current = head;
if (current == null) {
System.out.println("链表空 ");
return;
}
while (current!= null) {
System.out.print(current.iData + " ");
current = current.Child;
}
}
public void display(int i) {
Node current = head;
if (current == null) {
System.out.println("链表空 ");
return;
}
for(int j=0;j<i;j++)
{
System.out.print(current.iData+" ");
current=current.Child;
}
}
//构造链表
public SingleLink(int slong) {

for (int i = 0; i <slong; i++) {
Node nd = new Node(i);
if (head == null) {
head = nd;
} else {
Node current = head;
while (current.Child != null) {
current = current.Child;
}
current.Child = nd;
}
}
display();
}
//构造环，即尾部指向其中;
public void makeC(int data){
Node circle=new Node();
Node tail=new Node();
Node current=head;


if(current==null){System.out.println("链表为空，");return;}
//寻找环点；
while(current!=null)
{
if(current.iData==data){circle=current;System.out.print("已找到\n");break;}
current=current.Child;
}
System.out.println("找到为："+circle.iData);
current=head;
while(current.Child!=null)
{
current=current.Child;
}
tail=current;
//构建有环链表
tail.Child=circle;
display(15);

}
//判断是否有环;
//判断方法为，构建两个单链表不可能相等的结点A,B，不断移动，AB指向同一个结点，则有环。
//用=判断是否指向同一个，终止条件为B.child为null;
public boolean isc()
{
if(head==null||head.Child==null){System.out.println("链表为空或者那啥" );return false;}
Node currenta=head;
Node currentb=head.Child;
while(currentb.Child.Child!=null){
currentb=currentb.Child.Child;
currenta=currenta.Child;
if(currenta==currentb){System.out.println("\n链表有环" );return true;}
}
return true;
}
}

/**
*
* @author Administrator
*/
public class CSingleLink {

/**
* @param args the command line arguments 有环链表
*/
public static void main(String[] args) {
// TODO code application logic here
SingleLink sl = new SingleLink(10);
sl.makeC(4);
sl.isc();
}
}