Input:intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1Output:Reference of the node with value = 2Input Explanation:The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads...
Remove Duplicates from Sorted List 从链表中删除元素,一般需要两个指针来完成,分别记录要删除的链表元素和上一个元素。按照这个思路: # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def delete...
Input:intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2Output:No intersectionExplanation:From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skip...
begin to intersect at node c1. Notes: If the two linked lists have no intersection at all, returnnull. The linked lists must retain their original structure after the function returns. You may assume there are no cycles anywhere in the entire linked structure. Your code should preferably run...
If the two linked lists have no intersection at all, return null. The linked lists must retain their original structure after the function returns. You may assume there are no cycles anywhere in the entire linked structure. Your code should preferably run in O(n) time and use only O(1) ...
思路1 http://bookshadow.com/weblog/2014/12/04/leetcode-intersection-two-linked-lists/ 判断交点是否存在 如果两个链表有交点,则它们的最后一个节点一定是同一个节点。所以当pA/pB到达链表末尾时,分别记录下A和B的最后一个节点。如果两个链表的末尾节点不一致,说明两个链表没有交点。
输入:intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1输出:Intersected at '2'解释:相交节点的值为 2 (注意,如果两个链表相交则不能为 0)。 从各自的表头开始算起,链表 A 为 [0,9,1,2,4],链表 B 为 [3,2,4]。在 A 中,相交节点前有 3 个...
根据CYC刷题顺序开始,链表第一题, 视频播放量 24、弹幕量 0、点赞数 1、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 kodgv, 作者简介 ,相关视频:CYC链表3leetcode148 链表排序,CYC链表2leetcode206 反转链表,CYC树题2 leetcode543 二叉树直径,CYC树题1 leetcode1
Next challenges: (M) Linked List Cycle II 二.Linked List Cycle II Total Accepted: 61662 Total Submissions: 196038 Difficulty: Medium Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you ...
Linked List Cycle II 如果链表存在环,求出环的起点。 分析:见LeetCode中二分查找章节中Find the Duplicate Number的追逐法 privateListNodedetectCycle(ListNode head){ListNode fast=head;ListNode slow=head;while(fast!=null&&fast.next!=null){//遇到终点说明不是环fast=fast.next.next;slow=slow.next;if(fas...