示例1: 输入:intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3输出:Reference of the node with value = 8输入解释:相交节点的值为 8 (注意,如果两个列表相交则不能为 0)。从各自的表头开始算起,链表 A 为 [4,1,8,4,5],链表 B 为 [5
Leetcode 160. Intersection of Two Linked Lists 技术标签: Leetcode文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution Version 1 Version 2 Reference https://leetcode.com/problems/intersection-of-two-linked-lists/description/......
Can you solve this real interview question? Intersection of Two Linked Lists - Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, ...
Input: intersectVal =8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA =2, skipB =3Output: Reference of the nodewithvalue=8 Example 2: Input: intersectVal =2, listA = [0,9,1,2,4], listB = [3,2,4], skipA =3, skipB =1Output: Reference of the nodewithval...
LeetCode之Intersection of two linked list不同方法 AC完看答案发现答案超简单,而自己的方法有点过于复杂了,题目原意是找出两个链表第一个公共节点,如果没有则返回NULL。 看到题目后,我竟然想到可能存在交叉结构,结果通过反转一个链表来求出是否存在公共节点,但是却没法求出第一个公共节点,因此重新看回题目,发现...
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) memory. 记链表A的长度是lenA,最后一个结点为p;链表B的长度是lenB,最后...
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...
题目链接https://leetcode.com/problems/intersection-of-two-linked-lists/ 题意很好懂,问题在于如何找到相交的node,想到的最简单的方法从node的最后一个节点去往前数,遇到分叉,则返回当前节点,否则返回None。这里用 两个list去保存node,从list的最后一个位置开始往前进。 代码如下,提交后,超过 99.84% 提交代码的...
LeetCode 160. Intersection of Two Linked Lists,"题目"题意:寻找两个链表重合部分的起始点。题解:计算两个链表的长度,从到最后一个点距离相等的点,开始比较就可以了。
LeetCode_160. Intersection of Two Linked Lists 思路1:申请一个vector,用来存放其中一条链表的所有节点,当遍历另外一条链表时,如果该节点已经在vector中存放过了,说明找到了第一个相交的节点。 /** * Definition for singly-linked list. * struct ListNode {...