55链表中环的入口结点

题目描述

一个链表中包含环，请找出该链表的环的入口结点。


利用快慢指针
如果链表的环中有n个节点。快指针先走n步，然后同时走。当慢指针走的入口节点的时候，
快指针围绕着环走了一圈又回到了入口节点。

如何得到得到环中节点的个数？

利用快慢指针，找到环中的任意一节点，从该节点走一圈，并计数，当再次回到该节点的时候，便统计完了环中节点的个数。

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {
    public ListNode EntryNodeOfLoop(ListNode head)
        {
            ListNode pMeet = MeetingNode(head);
            if(pMeet==null) return null;

            ListNode fast = head.next;
               ListNode slow = head;
            for(ListNode p = pMeet.next;p!=pMeet;p=p.next)
                fast = fast.next;

            while(slow!=fast){
                slow = slow.next;
                fast = fast.next;

            }
            return fast;
    }



   private ListNode MeetingNode(ListNode head){
        if(head ==null|| head.next==null) return null;
        ListNode slow = head;
        ListNode fast = head.next.next;
        while(slow!=fast){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }




}

c++:20180729

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
};
*/
class Solution {
public:
    ListNode* EntryNodeOfLoop(ListNode* head)
    {
        
        ListNode* meetNode= meeting(head);
        if(meetNode==NULL) return NULL;
        ListNode* fast = head;
        ListNode* slow = head;
        for(ListNode* p = meetNode;p!=meetNode;p=p->next)        
            fast=fast->next;
        
        while(slow!=fast){
            fast = fast->next;
            slow = slow->next;
        }
        return fast;
    }

    ListNode* meeting(ListNode* head){
    if(head==NULL ||head->next==NULL) return NULL;
        ListNode* fast = head->next->next;
        ListNode* slow = head;
        while(slow!=fast){
            fast = fast->next->next;
            slow = slow->next;
        }
    return fast;
}
};