Linked List Cycle II

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

Note: Do not modify the linked list.

Solution

Solution using hash set.

/**
 * 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) {
        // 2
        Set<ListNode> set = new HashSet<>();

        ListNode current = head;
        while (current != null) {
            if (set.contains(current)) {
                return current;
            } else {
                set.add(current);
            }

            current = current.next;
        }

        return null;
    }
}

Other solution using two pointers.

public class Solution {
    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;
}

    public ListNode detectCycle(ListNode head) {
        if (head == null) {
            return null;
        }

        // If there is a cycle, the fast/slow pointers will intersect at some
        // node. Otherwise, there is no cycle, so we cannot find an entrance to
        // a cycle.
        ListNode intersect = getIntersect(head);
        if (intersect == null) {
            return null;
        }

        // To find the entrance to the cycle, we have two pointers traverse at
        // the same speed -- one from the front of the list, and the other from
        // the point of intersection.
        ListNode ptr1 = head;
        ListNode ptr2 = intersect;
        while (ptr1 != ptr2) {
            ptr1 = ptr1.next;
            ptr2 = ptr2.next;
        }

        return ptr1;
    }
}