142. Linked List Cycle II (Medium)


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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.


Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.


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


public class Solution {
    // |----x1-----|--p-x2---|
    // x1 represent the normal section of the list without circle, x2 represents the circle section of the list
    // Assume step length of slow is 1, fast encounters slow after slow walks S steps.
    // We can derive that the length slow walked is S, and fast is 2*S
    // Suppose fast and slow meets at point p, subject to (2*S-x1)%x2 = (S-x1)%x2=p-x1
    // A simplified formula of above is S%x2=0

    // Then slow continues to proceed forward x1 steps, the finally position is x1+(S+x1-x1)%x2=?
    // The simplified equation is x1+S%x2==x1+0=x1, that explains why after walking x1 steps(x1 length)
    // slow2 meets slow at the start of the circle
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;

        while (fast != null && fast.next != null) {
            slow = slow.next;

            // even step size
            fast = fast.next.next;

            if (fast != slow) {

            // fast == slow means both pointer fast and slow already run into a circle
            // and the pointer fast and slow can be any position on the circle

            ListNode slow2 = head;
            while (slow2 != slow) {
                slow = slow.next;
                slow2 = slow2.next;

            return slow;

        return null;

Incorrect Solutions


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