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?

 

Floyd's cycle detection algorithm, aka Tortoise and Hare Algorithm

ref: https://leetcode.com/problems/linked-list-cycle-ii/discuss/44793/O(n)-solution-by-using-two-pointers-without-change-anything

ref: http://fisherlei.blogspot.com/2013/11/leetcode-linked-list-cycle-ii-solution.html

 

假设cycle前的长度为X，cycle的长度为Y，快慢指针分别走过 2t 和 t 的路程后，相遇在K点，那么：

             慢指针       t = X + nY + K        (1）

             快指针     2t = X + mY+ K        (2)          m,n为未知数

(1) 代入到 (2) 中, 有 2X + 2nY + 2K = X + mY + K  =>   X+K  =  (m-2n)Y    (3)

这说明，X和K的关系是基于Y互补的，即两个指针相遇以后，再往下走X步就回到Cycle的起点了

 

time: O(n), space: O(1)

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