src/main/java/com/fishercoder/solutions/_142.java

package com.fishercoder.solutions;

import com.fishercoder.common.classes.ListNode;

import java.util.HashSet;
import java.util.Set;

public class _142 {

    public static class Solution1 {
        public ListNode detectCycle(ListNode head) {
            Set<ListNode> seen = new HashSet<>();
            while (head != null) {
                if (!seen.add(head)) {
                    return head;
                }
                head = head.next;
            }
            return null;
        }
    }

    public static class Solution2 {
        /**
         * This comment explains it really well for this solution:
         * https://leetcode.com/problems/linked-list-cycle-ii/discuss/44774/Java-O(1)-space-solution-with-detailed-explanation./44281
         *
         * When fast and slow meet for the first time at point P, fast travelled (a + b + c + b)
         * and slow travelled (a + b), and we know fast travels twice fast as slow, so we have:
         * a + b + c + b = 2*(a + b), this gives us a == c;
         * so at point P, we start a new slow2 pointer from the head, when both slow and slow2 travelled distance a, they must meet
         * at cycle entrance point Q.
         */
        public ListNode detectCycle(ListNode head) {
            ListNode slow = head;
            ListNode fast = head;
            while (fast != null && fast.next != null) {
                slow = slow.next;
                fast = fast.next.next;
                if (slow == fast) {
                    ListNode slow2 = head;
                    while (slow2 != slow) {
                        slow = slow.next;
                        slow2 = slow2.next;
                    }
                    return slow;
                }
            }
            return null;
        }
    }
}