LeetCode142 环形链表II

题目

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, 10?]. 
 -10? <= Node.val <= 10? 
 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

双指针法

从快慢指针相遇的节点和head节点同时走直到相遇即为环的入口
假设环以外的长度为a,相遇点走过的为b,未走过的为c，则环长为b+c
a+n(b+c)+b = 2(a+b) 可得a=(n-1)(b+c)+c;因此再次相遇点一定是入口

• 时间复杂度：O(n),n为链表的节点数
• 空间复杂度：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;
    }
}