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(); } } }