LeetCode [142] Linked List Cycle II

 142. Linked List Cycle II

Medium

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

 

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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62

//C++: 12ms /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *detectCycle(ListNode *head) { ListNode *p1 = head; ListNode *p2 = head; while(true){ if(!p2||!p2->next) return NULL; p1 = p1->next; p2 = p2->next->next; if(p1==p2) break; } p1 = head; while(p1!=p2){ p1 = p1->next; p2 = p2->next; } return p1; } }; //Java /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode detectCycle(ListNode head) { if (head==null || head.next==null) return null; ListNode s = head, f = head; while(s!=null && f!=null && f.next!=null){ s = s.next; f = f.next.next; if(s==f) break; } if(s!=f) return null; s = head; while(s!=f){ s = s.next; f = f.next; } return s; } }