题目意思:链表有环,返回true,否则返回false 思路:两个指针,一快一慢,能相遇则有环,为空了没环 ps:很多链表的题目:都可以采用这种思路 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCyc…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in t…

141. 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? 利用快慢指针,如果相遇则证明有环 注意边界条件: 如果只有一个node. public class Solution { public boolean hasCycle(ListNode head) { if(head==null |…

题意:判断链表是否有环. 分析:快慢指针. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { ListNode* fast = head; ListNode…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题意: 给定一个链表,判断是否有环 思路: 快慢指针 若有环,则快慢指针一定会在某个节点相遇(此处省略证明) 代码: public class Solution { public boolean hasCycle(ListNode head) { ListNode fast =…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断链表中是否有环,不能用额外的空间,可以使用快慢指针,慢指针一次走一步,快指针一次走两步,若是有环则快慢指针会相遇,若是fast->next==NULL则没有环. 值得注意的是:在链表的题中,快慢指针的使用频率还是很高,值得注意. /** * Definition for si…

Given a linked list, determine if it has a cycle in it. 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…

2.6 Given a circular linked list, implement an algorithm which returns the node at the beginning of the loop.DEFINITIONCircular linked list: A (corrupt) linked list in which a node's next pointer points to an earlier node, so as to make a loop in the…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

题目: Given a linked list, determine if it has a cycle in it. 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…