PAT甲级【2019年9月考题】——A1162 MergingLinkedLists【25】

7-2 Merging Linked Lists (25 分)

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

【声明】

　　由于此题还未上PAT官网题库，故没有测试集，仅仅是通过了样例，若发现错误，感谢留言指正。

Solution：

　　很简单，就是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。

 1 #include <iostream>
 2 #include <vector>
 3 #include <unordered_map>
 4 using namespace std;
 5 struct Node
 6 {
 7     int addr, val, next;
 8 }nodes[100010];
 9 int main()
10 {
11     int head, head1, head2, n;
12     cin >> head1 >> head2 >> n;
13     vector<Node>v1, v2, res;//存储列表值
14     for (int i = 0; i < n; ++i)
15     {
16         int a, b, c;
17         cin >> a >> b >> c;
18         nodes[a] = { a,b,c };
19     }
20     for (int p = head1; p != -1; p = nodes[p].next)
21         v1.push_back(nodes[p]);
22     for (int p = head2; p != -1; p = nodes[p].next)
23         v2.push_back(nodes[p]);
24     if (v1.size() > v2.size())
25         head = head1;
26     else
27     {
28         v2 = v1;//v2是短边
29         head = head2;
30     }
31     int k = 0;
32     while (head != -1)
33     {
34         res.push_back(nodes[head]);
35         ++k;
36         if (k % 2 == 0 && !v2.empty())//两个中间插一个
37         {
38             res.push_back(v2.back());
39             v2.pop_back();
40         }
41         head = nodes[head].next;
42     }
43     for (int i = 0; i < res.size() - 1; ++i)
44         printf("%05d %d %05d\n", res[i].addr, res[i].val, res[i+1].addr);
45     printf("%05d %d %d\n", res.back().addr, res.back().val, -1);
46     return 0;
47 }