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59 lines
1.9 KiB
Markdown
59 lines
1.9 KiB
Markdown
# [142. Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)
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## 题目
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Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
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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.
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Note: Do not modify the linked list.
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Example 1:
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```c
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Input: head = [3,2,0,-4], pos = 1
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Output: tail connects to node index 1
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Explanation: There is a cycle in the linked list, where tail connects to the second node.
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```
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Example 2:
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```c
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Input: head = [1,2], pos = 0
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Output: tail connects to node index 0
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Explanation: There is a cycle in the linked list, where tail connects to the first node.
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```
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Example 3:
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```c
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Input: head = [1], pos = -1
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Output: no cycle
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Explanation: There is no cycle in the linked list.
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```
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## 题目大意
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判断链表是否有环，不能使用额外的空间。如果有环，输出环的起点指针，如果没有环，则输出空。
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## 解题思路
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这道题是第 141 题的加强版。在判断是否有环的基础上，还需要输出环的第一个点。
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分析一下判断环的原理。fast 指针一次都 2 步，slow 指针一次走 1 步。令链表 head 到环的一个点需要 x1 步，从环的第一个点到相遇点需要 x2 步，从环中相遇点回到环的第一个点需要 x3 步。那么环的总长度是 x2 + x3 步。
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fast 和 slow 会相遇，说明他们走的时间是相同的，可以知道他们走的路程有以下的关系：
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```c
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fast 的 t = (x1 + x2 + x3 + x2) / 2
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slow 的 t = (x1 + x2) / 1
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x1 + x2 + x3 + x2 = 2 * (x1 + x2)
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所以 x1 = x3
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```
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所以 2 个指针相遇以后，如果 slow 继续往前走，fast 指针回到起点 head，两者都每次走一步，那么必定会在环的起点相遇，相遇以后输出这个点即是结果。
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