1.题目求两个链表的公共节点。
2.直接建立两个100001长度的数组，一个用来当作链表使用，一个用来当作后面的哈希查询使用。
2.遍历第一个链表，把第一个链表中存在的节点记录在哈希表中。
3.遍历第二个链表，如果哈希中存在某个节点的值，就break，最终输出该节点值。
4.把地址改为int存储，最后输出注意-1和其他的补0的情况。
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 1Input 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}, andNext 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
AC代码：
[c language=”++”]
//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
/*
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
11111 22222 9
00012 i 00002
00010 a 12345
00003 g -1
12345 D 00012
00002 n 00003
22222 B 23456
11111 L 00001
23456 e 00012
00001 o 00010
*/
int main(void)
{
int *m = new int[100001];
int *exist = new int[100001];
memset(m, -1, sizeof(m));
memset(exist, -1, sizeof(exist));
int sum;
int a, b;
scanf("%d %d %d", &a, &b, &sum);
m[a] = -1;
m[b] = -1;
for (int i = 0; i < sum; i++)
{
char tmp[10];
int pre, next;
scanf("%d %s %d", &pre, tmp, &next);
m[pre] = next;
}
int head = a;
while (head != -1)
{//把第一个链表存进哈希表 exist中
exist[head] = 1;
head = m[head];
}
head = b;
while (head != -1)
{
if (exist[head] == 1)
{
break;
}
head = m[head];
}
if (head != -1)
printf("%05d\n", head);
else
printf("%d\n", head);
return 0;
}
[/c]