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Given the head
of 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 (0-indexed). It is -1
if there is no cycle. Note that pos
is not passed as a parameter.
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.
Constraints:
[0, 104]
.-105 <= Node.val <= 105
pos
is -1
or a valid index in the linked-list.class Solution(object):
def detectCycle(self, head):
slow = fast = head
while fast and fast.next:
slow, fast = slow.next, fast.next.next
if slow == fast: break
else: return None # if not (fast and fast.next): return None
while head != slow:
head, slow = head.next, slow.next
return head
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) break;
}
if (!(fast && fast->next)) return NULL;
while (head != slow) {
head = head->next;
slow = slow->next;
}
return head;
}
};
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) break;
}
if (fast == null || fast.next == null) return null;
while (head != slow) {
head = head.next;
slow = slow.next;
}
return head;
}
}
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