Linked List Cycle (E)

Given a linked list, determine if it has a cycle in it.

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.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
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: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

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

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

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题意

判断给定链表中是否存在环。

思路

比较简单的就是将所有遍历到的结点记录下来，如果记录了两次则说明存在环。

$O(1)$O(1)空间的方法是使用快慢指针，慢指针每次走一步，快指针每次走两步，如果快指针追上慢指针则说明存在环。实际上可以看做一个相遇问题，如果快慢指针都在一个环中，且快指针距离慢指针还有n个结点的距离，那么经过n个回合后快指针必定与慢指针重合。

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代码实现 - Hash

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null) {
            return false;
        }

        Set<ListNode> set = new HashSet<>();
        while (head != null) {
            if (!set.add(head)) {
                return true;
            }
            head = head.next;
        }
        
        return false;
    }
}

代码实现 - 快慢指针

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null) {
            return false;
        }

        ListNode slow = head, fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (fast == slow) {
                return true;
            }
        }

        return false;
    }
}