Linked List Cycle II

Introduction Link to heading

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: Link to heading

Can you solve it without using extra space?

Solution Link to heading

Simple solution:

type ListNode struct {
    Val  int
    Next *ListNode
}
func detectCycle(head *ListNode) *ListNode {

    if head == nil || head == head.Next {
        return head
    }

    for i, j := one(head), two(head); i != nil && j != nil; i, j = one(i), two(j) {
       if i == j {
           meet := i
           if meet == head {
               return meet
           }
           for i, j := one(meet), one(head); i != nil && j != nil; i, j = one(i), one(j) {
               if i == j {
                   return i
               }
           }  
       }
    }

    return nil

}

func one(node *ListNode) *ListNode {
    return node.Next
}

func two(node *ListNode) *ListNode {
    next := node.Next
    if next == nil {
        return nil
    } else {
        return next.Next
    }
}

Explanation Link to heading

First i pointer goes one step a time, whereas second j pointer goes two steps a time. When they meet, first i pointer will ahead in T+X steps, whereas seconds pointer j will go ahead on N*2 steps. Second pointer j will definitely do one loop before meeting i, because it is faster, so it will go N+M steps ahead. At the same time j do two steps in one iteration, therefore N+M, must divide on 2, that gives M=N and N*2 steps for j. We know that N=T+X, whereas N is the loop size, T is the tail size and X is unknown position where pointers will meet. By using simple calculous we can find that T=N-X To find the T in loop, we need to continue going till the end of the loop using pointer i, that will give us N-X steps, and start j pointer from the head, since it will go exactly T steps, they will definitely meet with each other.