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?
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def detectCycle(self, head: ListNode) -> ListNode:
slow = fast = head
has_cycle = False
while fast and fast.next:
slow, fast = slow.next, fast.next.next
if slow == fast:
has_cycle = True
break
if not has_cycle:
return None
p = head
while p != slow:
p, slow = p.next, slow.next
return p/**
* 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) {
ListNode slow = head, fast = head;
boolean hasCycle = false;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
hasCycle = true;
break;
}
}
if (!hasCycle) {
return null;
}
ListNode p = head;
while (p != slow) {
p = p.next;
slow = slow.next;
}
return p;
}
}