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)