Given a linked list, return the node where the cycle begins. If there is no cycle, return null
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此题和之前的find the duplicate number比较像,代码如下:
1 /** 2 * Definition for singly-linked list. 3 * class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode detectCycle(ListNode head) { 14 ListNode fast = head; 15 ListNode slow = head; 16 while(fast!=null&&fast.next!=null){ 17 fast = fast.next.next; 18 slow = slow.next; 19 if(fast==slow){ 20 fast = head; 21 while(fast!=slow){ 22 fast = fast.next; 23 slow = slow.next; 24 } 25 return fast; 26 } 27 } 28 return null; 29 } 30 }