142. Linked List Cycle II
MediumLeetCodeGiven 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
How to solve the problem
- There is a fact we need to know in advance: when slow mmeets fast pointer, head meets slow at the node where the cycle begins.
Code
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = head
fast = head
while fast and fast.next: # when slow and fast.next is not None
slow = slow.next # slow move 1 step
fast = fast.next.next # fast move 2 step
if slow is fast: # when slow and fast meet each other
while slow is not head: # if index1(start from head) and slow(move 1 step everytime) is not meet, then keep moving
slow = slow.next
head = head.next
return slow
return None
Complexity
- Time complexity: O(n)
- Space complexity: O(1)
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