题目如下：
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.
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 N (<= 105), where the two addresses are the addresses of the first nodes of the two words, and N is the total number of nodes.
The address of a node is a 5-digit positive 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 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:
67890
Sample Input 2:
00001 00002 4 00001 a 10001 10001 s -1 00002 a 10002 10002 t -1
Sample Output 2:
-1
这道题目的关键词是suffix，suffix是后缀、词尾的意思，因此这道题目的难度有所降低，可以通过从尾部向前检索的方法来找到共同部分，但是题目所给的显然是单向链表结构，难以倒着查找，如果使用此方法还需要反转链表，因此我们可以考虑另一种方法。
STL中有红黑树容器set，我们首先建立一个容纳int的set，然后遍历第一个单词的链表，把所有结点的地址压入set，然后在对第二个链表遍历时，不断的把当前结点的地址从set中进行查找，如果找到了，说明到达了公共部分，直接输出此地址，一直没找到，则说明无公共部分。
对于题目中给出的这种用数字表示地址的链表结构，可以通过结构体数组实现，数组元素的索引即为自己的地址，元素为一个结构体，可以存储结点信息。
这个题目要注意在地址输出时如果地址位数不足5位要补充前导0。
具体代码如下：
// 注意前导0
// 巧用set来处理查找问题
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <set>
#define MAX 100001
struct Node{
char c;
int index;
int next;
};
Node nodes[MAX];
using namespace std;
int main()
{
int head1, head2, N;
cin >> head1 >> head2 >> N;
int myadd,nextadd;
char charset;
map<char,int> cmap;
for(int i = 0; i < N; i++){
scanf("%d %c %d",&myadd,&charset,&nextadd);
nodes[myadd].index = myadd;
nodes[myadd].c = charset;
nodes[myadd].next = nextadd;
cmap[charset] = myadd;
}
int cur = head1;
set<int> lists;
while(cur != -1){
int address = nodes[cur].index;
lists.insert(address);
cur = nodes[cur].next;
}
cur = head2;
int common = -1;
while(cur != -1){
int address = nodes[cur].index;
set<int>::iterator it = lists.find(address);
if(it != lists.end()){
common = address;
break;
}
cur = nodes[cur].next;
}
if(common == -1){
cout << common << endl;
}else{
printf("%05d\n",common);
}
return 0;
}