这就清晰了,X和K的关系是基于Y互补的。等于说,两个指针相遇以后,再往下走X步就回到Cycle的起点了。这就可以有O(n)的实现了。 from : http://fisherlei.blogspot.tw/2013/11/leetcode-linked-list-cycle-ii-solution.html Code : /** * Definition for singly-linked list. * struct ListNode { * int ...
1题目 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it...
Input:head = [3,2,0,-4], pos = 1Output:tail connects to node index 1Explanation:There is a cycle in the linked list, where tail connects to the second node. Example 2: Input:head = [1,2], pos = 0Output:tail connects to node index 0Explanation:There is a cycle in the linked ...
next = ListNode(list[i]) cur = cur.next return head ## 头元素指代一个链表 这里暂时还没想明白怎么初始化带有环的链表,所以这次就先直接在LeetCode中提交 solution。 class Solution: def hasCycle(self, head:ListNode) -> bool: seen = set() ## 空集合遍历收集元素 while head: ## 存在头结点 ...
next = pre; // 由于已是最后一次插入,所以无需再移动尾结点 } // 返回结果链表的头结点 head_pre.next } } 题目链接: Linked List Cycle : leetcode.com/problems/l 环形链表: leetcode-cn.com/problem LeetCode 日更第 53 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
利用LeetCode: 141. Linked List Cycle 题解 的快慢指针找到距离起点 n 个周期的节点(设慢指针移动 a+b 各节点, 则快指针移动 a+b+nT, 而快指针速度是慢指针的二倍,因此 2(a+b)=a+b+nT, 即 a...
leetcode -- Linked List Cycle -- 重点 AI检测代码解析 class Solution(object): def hasCycle(self, head): """ :type head: ListNode :rtype: bool """ if not head : return False slow, fast = head, head while fast and fast.next and fast != slow:#这里这个条件fast!=slow,使得在一开始...
Explanation: There is a cycle in the linked list, where tail connects to the second node. image.png 二、解决思路 方法一:使用HashMap存储遍历链表,并判重,O(n) 方法二:使用一快一慢指针检查是否相等,O(n) 三、算法实现 publicstatic booleanisCycle(Node head){if(head==null)returnfalse;boolean fla...
public class Solution { /** * @param head: The first node of linked list. * @return: The node where the cycle begins. * if there is no cycle, return null */ public ListNode detectCycle(ListNode head) { if(head == null || head.next == null) ...
Linked List Cycle I Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 快慢指针法 复杂度 时间O(N) 空间 O(1) 思路 这是一道非常经典的双指针题。我们从头设置一个快指针,一个慢指针。快指针一次走两步,慢指针一次走一步,如果快指...