142. Linked List Cycle II

• Difficulty: Medium

• Topics: Linked List, Two Pointers

• Similar Questions:

  • Linked List Cycle

  • Find the Duplicate Number

Problem:

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?

Solutions:

/**
 * 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* fast = head;
        ListNode* slow = head;

        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
            if (slow == fast) break; 
        }

        if (fast != slow || head == NULL || head->next == NULL)   return NULL;

        ListNode* another = head;
        while (another != slow) {
            another = another->next;
            slow = slow->next;
        }

        return another;
    }
};