原题链接：暂无
关键词：链表

Given two singly linked lists L 1 =a 1 →a 2 →…→a n−1 →a n L1=a1→a2→…→an−1→an and L 2 =b 1 →b 2 →…→b m−1 →b m L2=b1→b2→…→bm−1→bm . If n≥2m n≥2m , you are supposed to reverse and merge the shorter one into the longer one to obtain a list like a 1 →a 2 →b m →a 3 →a 4 →b m−1 … a1→a2→bm→a3→a4→bm−1… For example, given one list being 6→7 and the other one 1→2→3→4→5, you must output 1→2→7→3→4→6→5.

Input Specification：
Each input file contains one test case. For each case, the first line contains the two addresses of the first nodes of L 1 L1 and L 2 L2 , plus a positive N(≤10 5 ) N(≤105) which is the total number of nodes given. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next
where Address is the position of the node, Data is a positive integer no more than 10 5 105 , and Next is the position of the next node. It is guaranteed that no list is empty, and the longer list is at least twice as long as the shorter one.

Output Specification：
For each case, output in order the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input：

00100 01000 7
02233 2 34891
00100 6 00001
34891 3 10086
01000 1 02233
00033 5 -1
10086 4 00033
00001 7 -1

Sample Output：

01000 1 02233
02233 2 00001
00001 7 34891
34891 3 10086
10086 4 00100
00100 6 00033
00033 5 -1

题目大意： 给出两个链表，要求按照规则输出合并后的链表
思路： 先把节点数据存储下来，然后分别获取出两条链表的节点，然后进行按要求拼接。
代码：

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;const int N = 100000 + 10; 
struct Node
{int address, data, next;
}nodes[N];int main()
{int head, add1, add2, n;cin >> add1 >> add2 >> n;	//2个开始地址 结点数 vector<Node>v1, v2, res;	//存储链表值for (int i = 0; i < n; ++i){int a, b, c;cin >> a >> b >> c;nodes[address] = { a,b,c };}for (int p = add1; p != -1; p = nodes[p].next)v1.push_back(nodes[p]);for (int p = add2; p != -1; p = nodes[p].next)v2.push_back(nodes[p]);if (v1.size() > v2.size())head = add1;else{v2 = v1;	//v2是短链表 head = head2;}int k = 0;while (head != -1){res.push_back(nodes[head]);++k;if (k % 2 == 0 && !v2.empty())//两个中间插一个{res.push_back(v2.back());v2.pop_back();}head = nodes[head].next;}for (int i = 0; i < res.size() - 1; ++i)printf("%05d %d %05d\n", res[i].addr, res[i].val, res[i+1].addr);printf("%05d %d %d\n", res.back().addr, res.back().val, -1);return 0;
}