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?
/**
* 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;
}
//假设环的起点是第m个节点，总长度是n
ListNode fast = head;
ListNode slow = head;
while(fast!=null && fast.next!=null){
fast = fast.next.next;
slow = slow.next;
if(fast == slow){
break;
}
}
if(fast==null || fast.next ==null){
return null;
}
//第一次相遇在k节点,此时慢指针只需走n-k到达开头
//此时快指针已经走了m+k步
fast = head;
while(fast!=slow){
slow = slow.next;
fast = fast.next;
}
return slow;
}
}