Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next
pointer. Internally, pos
is used to denote the index of the node that tail's next
pointer is connected to. Note that pos
is not passed as a parameter.
Notice that you should 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.
Constraints:
- The number of the nodes in the list is in the range
[0, 104]
. -105 <= Node.val <= 105
pos
is-1
or a valid index in the linked-list.
Follow up: Can you solve it using O(1)
(i.e. constant) memory?
# 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 not has_cycle and fast and fast.next:
slow, fast = slow.next, fast.next.next
has_cycle = slow == fast
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 (!hasCycle && fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
hasCycle = slow == fast;
}
if (!hasCycle) {
return null;
}
ListNode p = head;
while (p != slow) {
p = p.next;
slow = slow.next;
}
return p;
}
}
/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function detectCycle(head: ListNode | null): ListNode | null {
let slow = head,
fast = head;
while (fast) {
slow = slow.next;
if (!fast.next) return null;
fast = fast.next.next;
if (fast == slow) {
let cur = head;
while (cur != slow) {
slow = slow.next;
cur = cur.next;
}
return cur;
}
}
return null;
}
/**
* 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* slow = head;
ListNode* fast = head;
bool hasCycle = false;
while (!hasCycle && fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
hasCycle = slow == fast;
}
if (!hasCycle) {
return nullptr;
}
ListNode* p = head;
while (p != slow) {
p = p->next;
slow = slow->next;
}
return p;
}
};
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var detectCycle = function (head) {
let slow = head;
let fast = head;
let hasCycle = false;
while (!hasCycle && fast && fast.next) {
slow = slow.next;
fast = fast.next.next;
hasCycle = slow == fast;
}
if (!hasCycle) {
return null;
}
let p = head;
while (p != slow) {
p = p.next;
slow = slow.next;
}
return p;
};
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func detectCycle(head *ListNode) *ListNode {
slow, fast := head, head
hasCycle := false
for !hasCycle && fast != nil && fast.Next != nil {
slow, fast = slow.Next, fast.Next.Next
hasCycle = slow == fast
}
if !hasCycle {
return nil
}
p := head
for p != slow {
p, slow = p.Next, slow.Next
}
return p
}