Given a linked list, return the node where the cycle begins. If there is no cycle, return null
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Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
141题的延伸,求出循环点。
可以用数学方法证明出slow与find相遇的位置一定是所求的点。
/** * 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 fast = head; ListNode slow = head; while( fast != null && fast.next != null ){ slow = slow.next; fast = fast.next.next; if( fast == slow ){ ListNode find = head; while( find != slow ){ find = find.next; slow = slow.next; } return find; } } return null; } }