给定一单链表的头指针head
1、判断链表是否又环。
2、如果有环，求环长以及环的起始节点。
package aglist;
class Node{
public final int value;
public Node next = null;
Node(int value){
this.value = value;
}
public static void addTail(Node head,Node node){
Node temp = head;
while(temp.next != null){
temp = temp.next;
}
temp.next = node;
}
public static Node getNode(Node head,int index){
int i = 0;
Node temp = head;
while(i < index && temp.next != null){
temp = temp.next;
i++;
}
return temp;
}
public static Node getTail(Node head){
Node temp = head;
while(temp.next != null){
temp = temp.next;
}
return temp;
}
}
public class Circle {
public static boolean hasCircle(Node head){
Node fast = head,slow = head;//两指针，一个步长为1，一个步长为2
while(fast.next != null && fast.next.next != null){
slow = slow.next;
fast = fast.next.next;
if(fast == slow){return true;}//两指针相遇，必有环，且相遇点在环内
}
return false;
}
private static Node getMeetNode(Node head){
Node fast = head,slow = head;
while(fast.next != null && fast.next.next != null){
slow = slow.next;
fast = fast.next.next;
if(fast == slow){break;}
}
return fast;
}
private static int getLengthOfCircle(Node meetNode){
int length = 1;
Node temp = meetNode.next;
while(temp != meetNode){//由于相遇点必在环内，只要顺着环走一圈，即可获得环的长度
length++;
temp = temp.next;
}
return length;
}
/*
* 在相遇点将环分开，视为形成两个单链表
* 一个单链表以head为头
* 另一个单链表以meetNode为头
* 该问题进而规约为求两个单链表的交点，交点即为环的起点
* 并且其交点到两链表头等长
* */
private static Node getStartOfCircle(Node head,Node meetNode){
while(head != meetNode){
head = head.next;
meetNode = meetNode.next;
}
return meetNode;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
try {
int[] ary = {3,4,5,7,1,40,10,50};
Node head = new Node(30);
for (int i = 0; i < ary.length; i++) {
Node.addTail(head, new Node(ary[i]));
}
Node.getTail(head).next = Node.getNode(head, 0);
if (hasCircle(head)) {
Node meetNode = getMeetNode(head);
int lenght = getLengthOfCircle(meetNode);
Node start = getStartOfCircle(head, meetNode);
System.out.println("length:"+lenght);
System.out.println("StartNode:"+start.value);
}
} catch (Exception e) {
// TODO: handle exception
e.printStackTrace();
}
}
}
package aglist;
class Node{public final int value;public Node next = null;Node(int value){this.value = value;}public static void addTail(Node head,Node node){Node temp = head;while(temp.next != null){temp = temp.next;}temp.next = node;}public static Node getNode(Node head,int index){int i = 0;Node temp = head;while(i < index && temp.next != null){temp = temp.next;i++;}return temp;}public static Node getTail(Node head){Node temp = head;while(temp.next != null){temp = temp.next;}return temp;}}public class Circle {public static boolean hasCircle(Node head){Node fast = head,slow = head;//两指针，一个步长为1，一个步长为2while(fast.next != null && fast.next.next != null){slow = slow.next;fast = fast.next.next;if(fast == slow){return true;}//两指针相遇，必有环，且相遇点在环内}return false;}private static Node getMeetNode(Node head){Node fast = head,slow = head;while(fast.next != null && fast.next.next != null){slow = slow.next;fast = fast.next.next;if(fast == slow){break;}}return fast;}private static int getLengthOfCircle(Node meetNode){int length = 1;Node temp = meetNode.next;while(temp != meetNode){//由于相遇点必在环内，只要顺着环走一圈，即可获得环的长度length++;temp = temp.next;}return length;}/* * 在相遇点将环分开，视为形成两个单链表 * 一个单链表以head为头 * 另一个单链表以meetNode为头 * 该问题进而规约为求两个单链表的交点，交点即为环的起点 * 并且其交点到两链表头等长 * */private static Node getStartOfCircle(Node head,Node meetNode){while(head != meetNode){head = head.next;meetNode = meetNode.next;}return meetNode;}public static void main(String[] args) {// TODO Auto-generated method stubtry {int[] ary = {3,4,5,7,1,40,10,50};Node head = new Node(30);for (int i = 0; i < ary.length; i++) {Node.addTail(head, new Node(ary[i]));}Node.getTail(head).next = Node.getNode(head, 0);if (hasCircle(head)) {Node meetNode = getMeetNode(head);int lenght = getLengthOfCircle(meetNode);Node start = getStartOfCircle(head, meetNode);System.out.println("length:"+lenght);System.out.println("StartNode:"+start.value);}} catch (Exception e) {// TODO: handle exceptione.printStackTrace();}}
}