链表判环问题

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();
        }
    }

}

 

posted @ 2017-09-21 23:35  Qcer  阅读(210)  评论(0编辑  收藏  举报