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 con
第二步设两个指针,一个从相遇点出发,另一个从原始起点出发,二者都是一次一步,当二者相遇时,所在的点即为环的入口. 第二步要会从数学上证明: 设进入环之前的直线距离为d,二者相遇时,慢指针在环上走过的距离是x,快指针绕环k(注意k必然>=1)圈外加x, 环的周长是C,则有: d+x = (x+d+kC) / 2 =...
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. 翻译过来就是找到链表中的环,并返回环起点的节点。
Given a linkedlist,returnthe node where the cycle begins. If there is no cycle,returnnull. Note: Do not modify the linkedlist. Follow up: Can you solve it without using extra space? 题目分析:就是判断并找到一条单链表相交的起始节点,如果没有相交,返回 NULL 即可 。 解题思路:先来看下面的这幅...
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 ...
【Leetcode】Linked List Cycle II https://leetcode.com/problems/linked-list-cycle-ii/ 题目: null. Note: Follow up: Can you solve it without using extra space? 思路: 首先确定循环是否存在,若存在,根据 循环结点个数/结点相对移动次数 就会相遇的规律 得到循环结点个数,再从头开始遍历,相对移动速度为...
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...
Problem link: https://leetcode.com/problems/linked-list-cycle-ii/有任何错误欢迎指出,有任何问题欢迎留言,谢谢观看, 视频播放量 94、弹幕量 1、点赞数 1、投硬币枚数 2、收藏人数 1、转发人数 1, 视频作者 dddeng12, 作者简介 ,相关视频:51. N 皇后 (N-queens),重
slow 1 step 每次 //用 do whlie 可以避免边界 ListNode *detectCycle(ListNode *head) { //有可能无环 auto fast = head; auto slow = head; //拉开差距 do{ //边界2 if( fast && fast -> next) fast = fast -> next; else return nullptr; //没有环 slow = slow -> next; fast = fast...
141 Linked..判断链表 LinkList 是否带循环。Given a linked list, determine if it has a cycle in it.To represent a cycle in t