[LintCode] Linked List Cycle 单链表中的环

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, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 开始以为这道题只需要注意不使用额外空间即可,于是写了个时间复杂度为O(n^2)暴力搜索算法,如下: /** * Dumped, Time Limit Exceeded */ class Solution { public: bool hasCycle(ListNode *hea

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? 这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸,可参见我之前的一篇文章 (http://www.cnblogs.com/grandyang/p/4137187.html). 还是要设

1.判断单链表是否有环 使用两个slow, fast指针从头开始扫描链表.指针slow 每次走1步,指针fast每次走2步.如果存在环,则指针slow.fast会相遇:如果不存在环,指针fast遇到NULL退出. 就是所谓的追击相遇问题: 2.求有环单链表的环长 在环上相遇后,记录第一次相遇点为Pos,之后指针slow继续每次走1步,fast每次走2步.在下次相遇的时候fast比slow正好又多走了一圈,也就是多走的距离等于环长. 设从第一次相遇到第二次相遇,设slow走了len步,则fast走

题意:给一个单链表,判断其是否出现环! 思路:搞两个指针,每次,一个走两步,另一个走一步.若有环,他们会相遇,若无环,走两步的指针必定会先遇到NULL. 1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution

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: 1 public class Solution { 2 public boolean hasCycle(ListNode head) { 3 ListNode f

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 1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 1

class Solution { public: bool hasCycle(ListNode *head) { if(head == NULL) return false; if(head->next == NULL) return false; ListNode* p1 = head; ListNode* p2 = head; p1 = p1->next; p2 = p2->next->next; while(p2 != NULL && p2->next