思路:使用快慢指针,快指针每次走两步,慢指针每次走一步,如果有环,则一定会快慢指针指向同一结点: 假设环的长度为n,先让一个指针走n步,另一个再开始走,当他们指针指向同一结点时,该结点就是环入口点 (在快慢指针相遇之后,慢指针指向表头,快指针留在相遇点,二者以每次一步走直到相遇,该相遇点就是环入口结点): 找到环入口结点之后,从入口结点开始遍历,每次遍历长度加一,如果下个结点等于入口结点,则返回长度 class ListNode(object): def __init__(self, datav…

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…

题目 Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 分析 判断链表是否有环,采用快慢指针,如果相遇则表示有环 AC代码 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(i…

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? 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…

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. 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, return the node where the cycle begins. If there is no cycle, returnnull. Follow up:Can you solve it without using extra space? 题意:给定链表,若是有环,则返回环开始的节点,没有则返回NULL 思路:题目分两步走,第一.判断是否有环,第二若是有,找到环的起始点.关于第一点,可以参考之前的博客 Linked list cycle.…

判断给定的链表中是否有环.如果有环则返回true,否则返回false. 解题思路:设置两个指针,slow和fast,fast每次走两步,slow每次走一步,如果有环的话fast一定会追上slow,判断fast==slow或者fast.next==slow即可判断 class ListNode { int val; ListNode next; ListNode(int x) { val = x; next = null; } } public class test1 { public boole…

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 |…