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所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。
'''
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?
'''
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
# Approach one
# 1、快慢指针判断链表是否有环
# 2、如果有环，根据相交点找出入环节点
# 设慢指针走的路程为L，则快指针走的路程为2L；设头结点到入环点的长度为K，入环点到相交点的长度为M，环的长度为Q, 快指针一共绕环n圈。
# 则 L = K + M
# 2L = K + M + Q*n
# 根据上面两式，可以得到 K + M = n*Q
# 进而可以得到 K = (n - 1)*Q + (L - M)
# 也就是说，当快慢指针相遇后，慢指针回到头结点，快指针同时在相交点出发（但也以步长1），它们再次相遇时会刚好走到入环点。代码如下。
fast = slow = head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
if fast and fast == slow: # 不可以写成 slow.val == fast.val
slow = head
while slow != fast:
slow = slow.next
fast = fast.next
return slow
return None
所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。