[Leetcode] Linked list cycle II chain ring start position

Given a linked list, return the node where the cycle begins. If There is no cycle, return null .

Follow up:
Can you solve it without using extra space?

Hide TagsLinked List Pointers began to make two a point, using the Val of node as a judge, in fact, based on memory judgment. Find the starting position of the chain ring, this painting slowly find the law can: thinking:

1. Use the fast pointer, the slow pointer each time before the further, the quick pointer every two steps
2. If the quick and slow hands meet, then move the fast pointer from the tag with the linked header to the previous
3. When the fast and slow hands meet again is the ring start position.

This realization, the time quickly o (n), and the Space O (1)

#include <iostream>using namespacestd;/** * Definition for singly-linked list.*/structListNode {intVal; ListNode*Next; ListNode (intx): Val (x), Next (NULL) {}};classSolution { Public: ListNode*detectcycle (ListNode *head) {        if(Head==null)returnNULL; ListNode* fast=head,*slow=Head;  while(1){            if(fast->next!=null) fast=fast->Next; Else returnNULL; if(fast->next!=null) fast=fast->Next; Else returnNULL; Slow=slow->Next; if(Fast==slow) Break; } Fast=Head;  while(1){            if(Fast==slow)returnslow; Fast=fast->Next; Slow=slow->Next; }        returnNULL; }};intMain () {ListNode Node1 (1), Node2 (2), Node3 (3), Node4 (4), NODE5 (5); Node1.next=&Node2; Node2.next=&Node3; Node3.next=&Node4; Node4.next=&Node5; Node5.next=&Node1;    Solution Sol; ListNode*ret = Sol.detectcycle (&Node1); if(ret==null) cout<<"NULL"<<Endl; Elsecout<<ret->val<<Endl; return 0;}

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[Leetcode] Linked list cycle II chain ring start position