内容简介: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
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?
难度:medium
题目:给定一链表,返回环的起始结点。如果无环返回空。为了表示链表中的环,使用整数来表示起如结点位置即尾结点所指向的位置。如果位置为-1则表示无环。
注意:不要修改链表。
思路:快慢指针。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle II.
Memory Usage: 35.2 MB, less than 100.00% of Java online submissions for Linked List Cycle II.
/** * 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 (null == head) { return head; } ListNode slow = head, fast = head; do { slow = slow.next; fast = fast.next; fast = (fast != null) ? fast.next : fast; } while (fast != null && slow != fast); if (null == fast) { return null; } for (slow = head; slow != fast; slow = slow.next, fast = fast.next); return slow; } }
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[英] Tom Hughes-Croucher、[英] Mike Wilson / 郑达韡 / 人民邮电出版社 / 2013-2 / 39.00元
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