LeetCode 142 - Linked List Cycle II

Difficulty: medium

Problem Description

English (Linked List Cycle II)

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

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 (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

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: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

graph LR

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

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

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:

graph LR

node1((1))

Input: head = [1], pos = -1
Output: no cycle
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 II)

给定一个链表的头节点 head ，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。

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）。如果 pos 是 -1，则在该链表中没有环。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。

不允许修改 链表。

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：返回索引为 1 的链表节点
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：返回索引为 0 的链表节点
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：返回 null
解释：链表中没有环。

提示：

• 链表中节点的数目范围在范围 [0, 10^4] 内
• -10^5 <= Node.val <= 10^5
• pos 的值为 -1 或者链表中的一个有效索引

进阶： 你是否可以使用 O(1) 空间解决此题？

Solution

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

class Solution {
public:
    ListNode* detectCycle(ListNode* head) {
        ListNode* cycle_node = nodeInCycle(head);
        if (cycle_node == nullptr) {
            return nullptr;
        }

        ListNode* curr = head;
        while (curr != cycle_node) {
            curr = curr->next;
            cycle_node = cycle_node->next;
        }

        return curr;
    }

private:
    static ListNode* nodeInCycle(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 fast;
            }
        }

        return nullptr;
    }
};