Linked List Cycle II

Link: https://leetcode.com/problems/linked-list-cycle-ii/

Idea

If there's a cyle, we have the following findings:

Here 'a' means the distance from head to entry point of the cycle. Node with value 1 is the point where the slow and fast pointer first meet. Its distance to the entry point (node 0) is b, while 'c' is the distance to node 0 via the other path. Since we get a = b, if we have one pointer to start from 3, and the other pointer starts from 1, they will meet at node 0 if they both move one node at a time.

Code

public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        boolean hasCycle = false;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                hasCycle = true;
                break;
            }
        }
        
        if (!hasCycle) {
            return null;
        }
        
        slow = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        
        return slow;
    }
}

Refercnne: https://www.jiuzhang.com/problem/linked-list-cycle-ii/