Problem
Write a program to find the node at which the intersection of two singly linked lists begins.
ExampleThe following two linked lists:
A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3
begin to intersect at node c1.
Note原理: A.length + Intersection + B.length = B.length + Intersection + A.length,也就是说,走了相同的三段长度之后,正好一起回到Intersection的起点。
Solutionpublic class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode l1 = headA, l2 = headB; while (l1 != l2) { l1 = l1 == null ? headB: l1.next; l2 = l2 == null ? headA: l2.next; } return l1; } }
文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。
转载请注明本文地址:https://www.ucloud.cn/yun/65611.html
摘要:先想到的是,其实也可以,只是需要在遍历的时候,添加到数组中的数要掉,略微麻烦了一点。在里跑的时候,也要快一点。另一种类似做法的就快的多了。如果是找出所有包括重复的截距呢 Problem Given two arrays, write a function to compute their intersection. Notice Each element in the result m...
Problem You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1s digit is at the head of the list. Write a f...
摘要:月下半旬攻略道题,目前已攻略题。目前简单难度攻略已经到题,所以后面会调整自己,在刷算法与数据结构的同时,攻略中等难度的题目。 Create by jsliang on 2019-07-30 16:15:37 Recently revised in 2019-07-30 17:04:20 7 月下半旬攻略 45 道题,目前已攻略 100 题。 一 目录 不折腾的前端,和咸鱼有什么区别...
摘要:先考虑和有无空集,有则返回另一个。新建链表,指针将和较小的放在链表顶端,然后向后遍历,直到或之一为空。再将非空的链表放在后面。 Problem Merge two sorted (ascending) linked lists and return it as a new sorted list. The new sorted list should be made by splici...
摘要:题目解答非常聪明的解法,因为两个的长度不一样,所以它让两个指针通过两次循环,来把两个都扫一遍。因为公共的部分相同,所以当它们相遇的时候,就是。 题目:Write a program to find the node at which the intersection of two singly linked lists begins. For example, the followin...
阅读 2391·2021-11-19 09:40
阅读 3556·2021-10-12 10:12
阅读 1863·2021-09-22 15:04
阅读 2878·2021-09-02 09:53
阅读 739·2019-08-29 11:03
阅读 1099·2019-08-28 18:11
阅读 1707·2019-08-23 15:28
阅读 3553·2019-08-23 15:05