Description

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: 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.

Follow-up: Can you solve it without using extra space?

Solutions

1. Two Pointers

  主要在捋清楚关系,其实不需要知道具体快指针和慢指针绕着环转了多少圈,以及第二次相遇(第二次肯定在入口处,因为两个指针在第二次都是一次一步,如果在入口没有遇到,以后肯定不可能再遇到了)在环中的指针在环中又走了多少圈,这些可以不考虑,只要代码检查到有两次相遇了,第二次相遇点肯定就是入口!

# Time: O(n)
# Space: O(1)
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        if not head:
            return None
        fast, slow = head, head
        while fast and slow:
            if fast.next:
                fast = fast.next.next
            else:
                return None
            slow = slow.next
            if fast == slow:
                break
        
        if not fast or not slow:
            return None
        
        restart = head
        while restart != slow:
            restart = restart.next
            slow = slow.next
        return slow
# 16/16 cases passed (52 ms)
# Your runtime beats 81.56 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (15.9 MB)

References

  1. 142. Linked List Cycle II