PAT A1032-Sharing（链表）

PAT A1032-Sharing（链表）

题目

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 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

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:

67890

Sample Input 2:

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

Sample Output 2:

-1

解题思路

把所有节点先存储，在遍历第一个单词的链表时，在总的节点链表里将遍历到的点标记出来，之后再遍历第二个单词的链表，当遍历到有第一单词标记的点时，说明该点就是我们要找的点把地址打出来即可。

代码

#include <stdio.h>

using namespace std;

// 每个节点的结构
typedef struct {
    bool flag;
    char Data;
    int Next;
} NODE;

int main() {
    int fir, sec, N, addr;
    NODE table[100010];// 题目说明最多不会超过100000条数据
    scanf("%d %d %d", &fir, &sec, &N);
    
    for(int i = 0; i < N; i++){
        scanf("%d ", &addr);// 方便快速查找
        scanf("%c %d", &table[addr].Data, &table[addr].Next);
//        Table[addr].flag = false;
    }
    while(fir != -1){
        table[fir].flag = true;// 第一个单词遍历过的节点flag都标记为true
        fir = table[fir].Next;
    }
    while(sec != -1){
        if(table[sec].flag == true){// 说明第一个单词也遍历过这个节点，直接输出地址
            printf("%05d\n", sec);// 格式要求是输出五位
            return 0;
        }
        sec = table[sec].Next;
    }
    printf("%d\n", -1);
    return 0;
}

之前用了<map>来实现，但是时间花销比较大，因为map得到的hash表是有序的，这部分的排序会花销时间，所以如果没有对排序的需求，并且hash表范围能用数组表示的时候最好还是用数组吧。