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
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, ...
Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: begin to intersect at node c1. Example 1: Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB...
代码 classSolution{public:ListNode*reverseList(ListNode*head){if(!head||!head->next)returnhead;ListNode*node=head,*last=NULL;while(node){ListNode*tmp=node->next;node->next=last;last=node;node=tmp;}returnlast;}intgetLength(ListNode*head){ListNode*node=head;int count=0;while(node){count++;no...
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笔记:160. Intersection of Two Linked Lists 问题: Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: image.png begin to intersect at node c1. Notes:...
Explanation: The two lists do not intersect, so return null. 1. 2. 3. 4. Notes: If the two linked lists have no intersection at all, return null. The linked lists must retain their original structure after the function returns.
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...
解题思路: 根据下图,先给定一个空的链表newList,然后判断传入的链表head是不是空链表或者链表元素只有一个,如果是,直接返回就可以。如果不是,则对链表进行迭代,然后给一个临时变量temp存储head.next,然后改变head.next的指向newList,然后把head赋值给newList,接着让head等于临时变量temp...
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。