PAT Grade --A1032 Sharing

To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, loading and being are stored as showed in Figure 1.

Figure 1

You are supposed to find the starting position of the common suffix (e.g. the position of i in Figure 1).

Input Specification:

Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive

Then

Address Data Next

whereAddress is the position of the node, Data is the letter contained by this node which is an English letter chosen from { a-z, A-Z }, and Next is the position of the next node.

Output Specification:

For each case, simply output the 5-digit starting position of the common suffix. If the two words have no common suffix, output -1 instead.

Sample Input 1:

11111 22222 9
67890 i 00002
00010 a 12345
00003 g -1
12345 D 67890
00002 n 00003
22222 B 23456
11111 L 00001
23456 e 67890
00001 o 00010

Sample Output 1:

Sample Input 2:

00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 t -1

Sample Output 2:

-1

本以为，节点next出现两次，该节点就是公共节点，也包括-1在内，在牛客上测试通过，但在pat上测试失败，后来才感觉pat上可能出现公共节点在头结点上，故使用flag标记法，两条链表走过相同节点，则为公共节点

 1 #include <iostream>
 2 #include <unordered_map>
 3 using namespace std;
 4 int main()
 5 {
 6     int head1, head2, N, addr, next;
 7     char c;
 8     cin >> head1 >> head2 >> N;
 9     unordered_map<int, pair<int, bool>>list;
10     for (int i = 0; i < N; ++i)
11     {
12         cin >> addr >> c >> next;
13         list[addr] = make_pair(next,false);
14     }
15     for (int i = head1; i != -1; i = list[i].first)
16         list[i].second = true;
17     for (int i = head2; i != -1; i = list[i].first)
18     {
19         if (list[i].second == true)
20         {
21             printf("%05d\n", i);
22             return 0;
23         }
24     }
25     printf("-1\n");
26     return 0;
27 }