Leetcode-Linked List Cycle II

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?

Analysis:

Definitions:
Cycle = length of the cycle, if exists.
C is the beginning of Cycle, S is the distance of slow pointer from C when slow pointer meets fast pointer.

Distance(slow) = C + S, Distance(fast) = 2 * Distance(slow) = 2 * (C + S). To let slow poiner meets fast pointer, only if fast pointer run 1 cycle more than slow pointer. Distance(fast) - Distance(slow) = Cycle
=> 2 * (C + S) - (C + S) = Cycle
=> C + S = Cycle
=> C = Cycle - S
=> This means if slow pointer runs (Cycle - S) more, it will reaches C. So at this time, if there's another point2 running from head
=> After C distance, point2 will meet slow pointer at C, where is the beginning of the cycle.

Solution:

/**
 * 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, fast = head;
        while (fast != null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
            if (slow==fast){
                ListNode p2 = head;
                while (p2!=slow){
                    slow = slow.next;
                    p2 = p2.next;
                }
                return slow;
            }
        }
        return null;
    }
}