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142. Linked List Cycle II

2022-05-15 07:49:20Sterben_ Da

142. Linked List Cycle II

Medium

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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:

  • 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: Optional[ListNode]) -> Optional[ListNode]:
        """
         Their thinking ( Others' ):
         For the problem of finding loop in linked list , There is a general solution —— Speed pointer (Floyd  Circle method ). Given two pointers ,
         Named as  slow  and  fast, The starting position is at the beginning of the list . Every time  fast  Two steps forward ,slow  Take a step forward . If  fast
         You can go to the end , So there's no loop ; If  fast  You can go on forever , So there must be a loop , And must exist 
         In a moment  slow  and  fast  meet . When  slow  and  fast  The first time I met , We will  fast  Move back to the beginning of the list , and 
         Give Way  slow  and  fast  One step at a time . When  slow  and  fast  The second time we met , The node that meets is the starting point of the loop .
        """
        slow = fast = head
        cycle = False
        while fast:
            slow = slow.next
            fast = fast.next
            if fast is None:
                return None
            if not cycle:
                fast = fast.next
            if slow == fast:
                if cycle or fast == head:
                    #  Second touch or already at the beginning 
                    return slow
                cycle = True
                fast = head
                continue
        return None

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