技術標籤：資料結構連結串列

題目連結：https://pintia.cn/problem-sets/994805342720868352/problems/994805460652113920

1032 Sharing (25 分)

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

Figure 1

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

Input Specification:

Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positiveN(≤10​5​​), where the two addresses are the addresses of the first nodes of the two words, andNis the total number of nodes. The address of a node is a 5-digit positive integer, and NULL is represented by−1.

ThenNlines follow, each describes a node in the format:

Address Data Next

whereAddressis the position of the node,Datais the letter contained by this node which is an English letter chosen from { a-z, A-Z }, andNextis 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

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

題目大意：給定兩個連結串列，判斷這兩個連結串列是否有公共部分

解題思路：靜態連結串列方法，多定義一個flag變數，用來記錄flag是否在第一條連結串列中出現，是就是true，否false。從第一條連結串列首地址開始遍歷，經過的節點flag為true，列舉第二個節點，如果發現該節點flag值為true，說明是一個共用節點，如果沒有則輸出-1。不足5位的整數需要補0

程式碼：

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int ma = 100010;
struct node //靜態連結串列
{
    char data; //資料
    int next;  //下一個結點的地址
    bool flag; //標記
} a[ma];       //地址作為索引

int main()
{
    int n1, n2, n; //第一個連結串列和第二個連結串列的初始地址
    cin >> n1 >> n2 >> n;
    int index, next;
    char data;
    for (int i = 0; i < n; i++)
    {
        cin >> index;        //這個節點的地址
        cin >> data >> next; //資料及下一個節點的地址
        a[index] = {data, next, false};
    }
    for (int i = n1; i != -1; i = a[i].next)
    {
        a[i].flag = true; //第一個連結串列都標記為true
    }
    for (int i = n2; i != -1; i = a[i].next)
    {
        if (a[i].flag == true) //如果標記為true說明此節點是共同節點
        {
            printf("%05d", i);
            return 0;
        }
    }
    printf("-1");
    return 0;
}