The list of ring intersection problems

 1. Given a list, the list is determined whether there is a ring (as defined in the two references, the speed of the pointer, will eventually meet)

public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode fast=head;
        ListNode slow=head;
        do{
            if(fast==null)
            {
               return false;//空链表
            }
            fast=fast.next;
            if(fast==null)
                return false;//无环链表
            slow=slow.next;
            fast=fast.next;
        }while(fast!=slow);//知道他两相等，代表相遇，即有环
       return true;
        
    }
}

 

2. Given a list, the list starts to return into the first ring node. If chain acyclic, it is returned  null. (Define two references, a starting from scratch, starting from the meeting point to another, while walking, the first entry node is the point of encounter of the ring) 

public class Solution {
    public ListNode detectCycle(ListNode head) {
         ListNode fast=head;
        ListNode slow=head;
        do{
            if(fast==null)
            {
               return null;//空链表
            }
            fast=fast.next;
            if(fast==null)
                return null;//无环链表
            slow=slow.next;
            fast=fast.next;
        }while(fast!=slow);//知道他两相等，代表相遇，即有环
        ListNode p=head;//p从头出发
        ListNode q=slow;//q从相遇结点出发
        while(p!=q)
        {
            p=p.next;
            q=q.next;
        }
        return p;
        
    }
}

3. The sum of two intersection nodes of the list (the length of the two lists are determined, the definition of two reference points two lists, wherein a step difference go list of references, and then go together, if they are same, represents the intersection )

public class Solution {
    private int getLength(ListNode head){
        int len=0;
        for(ListNode cur=head;cur!=null;cur=cur.next)
        {
            len++;
        }
        return len;
    }//遍历求结点数
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int lena=getLength(headA);//链表a的结点数
        int lenb=getLength(headB);//链表b的结点数
        int diff=lena-lenb;//结点数差
        ListNode longer=headA;
        ListNode shorter=headB;
        if(lena<lenb)
        {
            longer=headB;
            shorter=headA;
            diff=lenb-lena;
        }
        for(int i=0;i<diff;i++)
            longer=longer.next;//结点数大的先走链表之差步。然后一起走，若相等则相交
        while(longer!=shorter)
        {
            longer=longer.next;
            shorter=shorter.next;
        }
        return longer;
    }
}