題目

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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

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.

 Constraints: 
 The number of the nodes in the list is in the range [0, 10⁴]. 
 -10⁵ <= Node.val <= 10⁵ 
 pos is -1 or a valid index in the linked-list. 

 Follow up: Can you solve it using O(1) (i.e. constant) memory? 
 

方法

雙指標法

從快慢指標相遇的節點和head節點同時走直到相遇即為環的入口
假設環以外的長度為a,相遇點走過的為b,未走過的為c，則環長為b+c
a+n(b+c)+b = 2(a+b) 可得a=(n-1)(b+c)+c;因此再次相遇點一定是入口

• 時間複雜度：O(n),n為連結串列的節點數
• 空間複雜度：O(1)

public class Solution {
    public ListNode detectCycle(ListNode head) {
        if(head==null){
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while(fast!=null){
            if(fast.next==null){
                return null;
            }else{
                fast = fast.next.next;
            }
            slow = slow.next;
            if(fast==slow){
                ListNode ptr = head;
                while(ptr!=slow){
                    ptr = ptr.next;
                    slow = slow.next;
                }
                return ptr;
            }
        }
        return null;
    }
}