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:
- 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?
Solution 1: Hashtset
Time complexity: O(n)
Space complexity: O(n)
C++
1 2 3 4 5 6 7 8 9 10 |
// Author: Huahua class Solution { public: ListNode *detectCycle(ListNode *head) { unordered_set<ListNode*> s; for (ListNode* cur = head; cur; cur = cur->next) if (!s.insert(cur).second) return cur; return nullptr; } }; |
Solution: Fast slow pointers
Time complexity: O(n)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 9 10 |
// Author: Huahua class Solution { public: ListNode *detectCycle(ListNode *head) { unordered_set<ListNode*> s; for (ListNode* cur = head; cur; cur = cur->next) if (!s.insert(cur).second) return cur; return nullptr; } }; |
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