List Cycle

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

Try solving it using constant additional space.

Example :

Input : 

                  ______
                 |     |
                 \/    |
        1 -> 2 -> 3 -> 4

Return the node corresponding to node 3.

Method:

Slow fast pointer

Solution:

Time: O(n)
Space: O(1)

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     public int val;
 *     public ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode a) {
        if (a == null || a.next == null) return null;
        ListNode slow = a.next;
        ListNode fast = a.next.next;
        while (slow != fast && fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        if (slow != fast) return null;
        slow = a;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}