142. Linked List Cycle II.md

142. Linked List Cycle II

Question

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

Thinking:

• Method 1: [如何求出链表的入环点]（https://github.com/Seanforfun/Algorithm/blob/master/leetcode/Offer/Question2_6.md）

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                ListNode cur = dummy;
                while(cur != slow){
                    cur = cur.next;
                    slow = slow.next;
                }
                return cur;
            }
        }
        return null;
    }
}

二刷

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy, fast = dummy;
        while(slow.next != null && fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast) break;
        }
        if(fast.next == null || fast.next.next == null) return null;
        slow = dummy;
        while(slow != fast){
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}

Third Time

• Method 1: two pointer

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