Leetcode 142. Linked List Cycle II
Linked List Cycle II - LeetCode
Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null. There is a…
head of a linked list, return the node where the cycle begins. If there is no cycle, return
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.
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.
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.
Input: head = , pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
- The number of the nodes in the list is in the range
-105 <= Node.val <= 105
-1or a valid index in the linked-list.
Follow up: Can you solve it using
O(1) (i.e. constant) memory?
- if you just want to check the ListNode had cycle, we can use two pointer, one move faster, another move slower. If they meet, it means the ListNode had a cycle.
- We want to know where they first meet. If we could track every position we have been to, we can have the answer.
- Trick part is we can start a new ListNode from the head, and move the slow pointer and new pointer together. The place they meet will be the answer.
- below photo from wmhlailenow help me understand the trick.
Faster pointer run a + b +c +b (total length = a + b + c)
Slower pointer run a + b
Faster pointer run 2 times faster than slower pointer; a + 2b + c = 2(a+b);
a = c