# 142.Linked-List-Cycle-II

## 142. Linked List Cycle II

## 题目地址

<https://leetcode.com/problems/linked-list-cycle-ii/>

## 题目描述

```
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?
```

## 代码

### Approach #1 Hash Table

```java
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
  public ListNode detectCycle(ListNode head) {
        Set<ListNode> visited = new HashSet<ListNode>();

    ListNode node = head;
    while (node != null) {
      if (visited.contains(node)) {
        return node;
      }
      visited.add(node);
      node = node.next;
    }

    return null;
  }
}
```

### Approach 2: Floyd's Tortoise and Hare

```java
public class Solution {
  public ListNode detectCycle(ListNode head) {
        if (head == null)  return null;

        ListNode intersect = getIntersect(head);
        if (intersect == null)     return null;

        ListNode ptr1 = head;
        ListNode ptr2 = intersect;
        while (ptr1 != ptr2) {
            ptr1 = ptr1.next;
            ptr2 = ptr2.next;
        }
        return ptr1;
    }

    private ListNode getIntersect(ListNode head) {
        ListNode tortoise = head;
        ListNode hare = head;

        // A fast pointer will either loop around a cycle and meet the slow
        // pointer or reach the `null` at the end of a non-cyclic list.
        while (hare != null && hare.next != null) {
            tortoise = tortoise.next;
            hare = hare.next.next;
            if (tortoise == hare) {
                return tortoise;
            }
        }

        return null;
    }
}
```


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