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0142-linked-list-cycle-ii.adoc

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142. Linked List Cycle II

{leetcode}/problems/linked-list-cycle-ii/[LeetCode - Linked List Cycle II^]

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.
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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.
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Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
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Follow-up:

Can you solve it without using extra space?

思路分析

这是 Floyd’s Tortoise and Hare (Cycle Detection) 算法。

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\begin{aligned} 2 \cdot \text { distance }(\text { tortoise }) &=\text { distance }(\text {hare}) \\ 2(F+a) &=F+a+b+a \\ 2 F+2 a &=F+2 a+b \\ F &=b \end{aligned}

这里解释一下：兔子跑的快，相遇时，兔子跑过的距离是乌龟跑过的距离的两倍。兔子沿着环，跑了一圈多，有多跑了 a 才相遇，所以距离是： \(F+a+b+a\)。

这个思路还可以用于解决 287. Find the Duplicate Number。

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一刷
link:{sourcedir}/_0142_LinkedListCycleII.java[role=include]
二刷
link:{sourcedir}/_0142_LinkedListCycleII_2.java[role=include]
三刷
link:{sourcedir}/_0142_LinkedListCycleII_3.java[role=include]

思考题

没想到有这么多环形探测算法！

尝试其他探测环形算法。