Linked List Cycle II | Leetcode

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

Follow up:
Can you solve it without using extra space?

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Solution: Maintain two pointers slow and fast, where slow advances by one node at a time, and fast advances by two nodes at a time. When they meet, reset slow to the head of the list. Then advance both slow and fast by one node at a time until they meet at the beginning of the cycle.

/**
 * 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) {
        ListNode slow = head;
        ListNode fast = head;
        do {
            if (fast == null || fast.next == null) return null;
            fast = fast.next.next;
            slow = slow.next;
        } while (fast != slow);
        
        slow = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        
        return slow;
    }
}