[] Wins the offer list entry node in the ring

Title link: the list entry ring node

 

Meaning of the questions: to a list, if containing ring, find the entry node ring of the list, otherwise the output null.

 

Solution: We set up two pointers, a fast walk, a slow you down. If there is a ring, the pointer will meet speed.

Further, a scratch two hands go, a meeting point go away, and finally an inlet in the ring will be met.

So let's find a number of links point, let go fast pointer c point of a link, you can find the node ring entrance.

 

Code:

 1 /*
 2 struct ListNode {
 3     int val;
 4     struct ListNode *next;
 5     ListNode(int x) :
 6         val(x), next(NULL) {
 7     }
 8 };
 9 */
10 class Solution {
11 public:
12     ListNode* EntryNodeOfLoop(ListNode* pHead){
13         //快指针和慢指针
14         ListNode* fast = pHead,*slow=pHead->next;
15         while(fast!=slow && fast!=NULL && slow != NULL){
16             slow = slow->next;
17             fast = fast->next;
18             if(fast!=NULL)    fast = fast->next;
19         }
20         //链表环节点个数
21         int cnt = 1;
22         ListNode *node = fast->next;
23         if(fast == slow && fast!=NULL){
24             while(node != fast){
25                 node = node->next;
26                 cnt++;
27             }
28         }
29         else    return NULL;
30        
31         fast = pHead;slow = pHead;
32         //fast先走cnt
33         for(int i = 0; i < cnt; i++)    fast = fast->next;
34         //找到入口
35         while(fast != slow){
36             fast = fast->next;
37             slow = slow->next;
38         }
39         return fast;
40     }
41 };