【PAT】【链表遍历】1032 Sharing (25 分)_1032 sharing最后一个测试点超时-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_52995942/article/details/118487299

该博客介绍了如何使用C++解决一道编程题，题目涉及字符串链表的构建和共享后缀的查找。作者首先分享了原始的解题思路，通过创建映射表记录节点出现次数来寻找共同后缀，但这种方法在处理大规模数据时效率较低。随后，作者参考了他人解题方案，优化了代码，通过一次遍历标记共享节点，大大提高了运行速度。博客还提供了两组样例输入和对应的输出结果。

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题目链接：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 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

// raw word:

// sublist子表 suffix后缀

// 解题思路：

// 求后缀中第一个元素的地址，如果没有后缀的话则输出-1

// 只需要对两个链表分别遍历一遍，重复遍历到的第一个元素则是答案

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
// 结点
struct node{
    string id, next;
    char c;
};
// 对结点地址进行映射，如果地址稀疏的的话会极大节省空间
// 但是这道题n的数量级是n^5，地址是五位数，所以完全没必要，反而增加了时间和空间复杂度
// 导致最后一个测试点接近超时
map<string, int> mp; 
// 用来计数，实际上在结构体中增加flag项是更优的做法
map<string, int> cnt;
const int maxn = 1e5 + 5;
node a[maxn];
int main(){
    // 读入数据
    string s1, s2; cin >> s1 >> s2;
    int n; cin >> n;
    string ans;
    for(int i = 1; i <= n; i++){
        cin >> a[i].id >> a[i].c >> a[i].next;
        mp[a[i].id] = i; // 地址映射
    }
    // 第一次遍历
    mp["-1"] = 0;
    int u = mp[s1];
    while(u){
        ++cnt[a[u].id];
        u = mp[a[u].next];
    }
    // 第二次遍历
    u = mp[s2];
    while(u){
        if(++cnt[a[u].id] > 1){
            cout << a[u].id << endl;
            return 0;
        }
        u = mp[a[u].next];
    }
    cout << -1 << endl;
    return 0;
}

// 参考了柳婼题解，优化后代码如下：

#include <iostream>
#include <cstdio>
using namespace std;

struct node{
    int next;
    char c;
    bool flag = 0;
};
const int maxn = 1e6 + 5;
node a[maxn];
int main(){
    int s1, s2, n; cin >> s1 >> s2 >> n;
    string ans;
    for(int id, i = 1; i <= n; i++){
        cin >> id;
        cin >> a[id].c >> a[id].next;
    }
    int u = s1;
    while(~u){
        a[u].flag = true;
        u = a[u].next;
    }
    u = s2;
    while(~u){
        if(a[u].flag){
            printf("%05d\n", u);
            return 0;
        }
        u = a[u].next;
    }
    cout << -1 << endl;
    return 0;
}