[勇者闯LeetCode] 142. Linked List Cycle II

[勇者闯LeetCode] 142. Linked List Cycle II

Description

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

null

.

Information

Tags: Linked List | Two Pointers

Difficulty: Medium

Solution

使用快慢指针fast和slow从链头开始扫描，fast每次走两步，slow每次走一步，当fast与slow相遇时说明有环，此时将fast重新从链头开始扫描，且将扫描步长改为一步，那么可证明，当fast和slow两次相遇时，它们所指向的就是环的起点。

Python Code

# 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
"""
fast, slow = head, head
while fast and fast.next:
fast, slow = fast.next.next, slow.next
if fast is slow:
fast = head
while fast is not slow:
fast, slow = fast.next, slow.next
return fast
return None