LeetCode 141 - Linked List Cycle

Difficulty: easy

Problem Description

English (Linked List Cycle)

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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

Example 1:

graph LR

node1((3))
node2((2))
node3((0))
node4(("-4"))

node1 ---> node2 ---> node3 ---> node4
node4 ---> node2

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

Example 2:

graph LR

node1((1))
node2((2))

node1 ---> node2
node2 ---> node1

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

Example 3:

graph LR

node1((1))

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

Constraints:

• The number of the nodes in the list is in the range [0, 10^4].
• -10^5 <= Node.val <= 10^5
• pos is -1 or a valid index in the linked-list.

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

Chinese (环形链表)

给你一个链表的头节点 head ，判断链表中是否有环。

如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，评测系统内部使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。注意：pos 不作为参数进行传递。 仅仅是为了标识链表的实际情况。

如果链表中存在环 ，则返回 true 。 否则，返回 false 。

示例 1：

graph LR

node1((3))
node2((2))
node3((0))
node4(("-4"))

node1 ---> node2 ---> node3 ---> node4
node4 ---> node2

输入：head = [3,2,0,-4], pos = 1
输出：true
解释：链表中有一个环，其尾部连接到第二个节点。

示例 2：

graph LR

node1((1))
node2((2))

node1 ---> node2
node2 ---> node1

输入：head = [1,2], pos = 0
输出：true
解释：链表中有一个环，其尾部连接到第一个节点。

示例 3：

graph LR

node1((1))

输入：head = [1], pos = -1
输出：false
解释：链表中没有环。

提示：

• 链表中节点的数目范围是 [0, 10^4]
• 10^5 <= Node.val <= 10^5
• pos 为 -1 或者链表中的一个 有效索引 。

进阶： 你能用 O(1)（即，常量）内存解决此问题吗？

Solution

/**
 * Definition for singly-linked list.
 */
struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    bool hasCycle(ListNode* head) {
        ListNode* fast = head;
        ListNode* slow = head;

        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;

            if (slow == fast) {
                return true;
            }
        }

        return false;
    }
};