Sword finger offer notes (14), Turing Java Architect vip video

[1, 2, 3, 4, 5, 6]

Note that 2 here represents a node numbered 2, and the node number starts from 0. Therefore, the node with number 2 is the node with val equal to 3.

Then the inlet node of the output ring is 3



**thinking**



**(Linked list, fast and slow pointer scanning)** O ( n ) O(n) O(n)  

The method of this question is ingenious.  

With two pointers f i r s t , s e c o n d first,second first,second Start from the starting point, f i r s t first first One step at a time, s e c o n d second second Take two steps at a time.  

If in the process s e c o n d second second get to`null`，Then there is no ring. Otherwise when f i r s t first first and s e c o n d second second After meeting, let f i r s t first first Return to the starting point, s e c o n d second second Stay where you are, and then the two pointers take one step at a time. When they meet, the meeting point is the entrance of the ring.



![](https://img-blog.csdnimg.cn/2021042619071423.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NTYyOTI4NQ==,size_16,color_FFFFFF,t_70)



Proof: as shown in the figure above, a Is the starting point, b It's the entrance to the ring, c Is the first meeting point of two pointers, ab The distance between is x，bc The distance between is y.   

Then when f i r s t first first get to b Due to s e c o n d second second than f i r s t first first Take twice as many roads, so s e c o n d second second Already from b Started walking on the ring x Step, may be more than 1 lap, distance b Still bad y Step (this is because the first meeting point is b after y Step, let's first return b Point, then s e c o n d second second Will retire 2 y Step, that is, distance b It's a little bad y Step); therefore s e c o n d second second from b Point go x+y Step back b A little, so s e c o n d second second from c Let's go. Let's go x You can just walk to b Point and let f i r s t first first Start from scratch, go x You can just walk there b Point. So the second meeting point is b Point.



**Time complexity**  

f i r s t first first A total of 2 x + y 2x+y 2x+y Step, s e c o n d second second A total of 2 x + 2 y + x 2x+2y+x 2x+2y+x Step, so the two pointers go a total of 5 x + 3 y 5x+3y 5x+3y Step. Because when the first time f i r s t first first get to b At 1:00, s e c o n d second second One lap at most to catch up f i r s t first first，therefore y Less than the length of the ring, so x + y x+y x+y Less than or equal to the total length of the linked list. So the total time complexity is O ( n ) O(n) O(n). 



**code**

/**

Definition for singly-linked list.
struct ListNode {
```
int val;
```
```
ListNode *next;
```

ListNode(int x) : val(x), next(NULL) {}

};

class Solution {

public:

ListNode *entryNodeOfLoop(ListNode *head) {

    if(!head || !head->next) return NULL;//If the first or second node of the linked list is empty

    ListNode *first = head, *second = head;

    while(first && second)

    {

        first = first->next;

        second = second->next;

        if(second) second = second->next;

        else return NULL;

        if(first == second)

        {

            first = head;

            while(first!=second)

            {

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