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Deduplication on a Linked List

题目链接：1097 Deduplication on a Linked List (25 point(s))

题干大意

链表的去重，绝对值相同即为重复

思路

设两个结果链表，分别存储，注意重复的链表不一定存在，要先判断是否为空才能输出

AC代码

C++
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#include <algorithm>
#include <iomanip>
#include <iostream>
#include <vector>
using namespace std;

const int MaxN = 1e6 + 10;
const int MaxK = 1e4 + 10;
struct node {
    int data{};
    int next{};
} l[MaxN];
int status[MaxK]{0};

int main()
{
    int         N, first;
    vector<int> res1, res2;
    cin >> first >> N;
    int address;
    for (int i = 0; i < N; ++i) {
        cin >> address >> l[address].data >> l[address].next;
    }
    while (first != -1) {
        if (status[abs(l[first].data)] == 0) {
            res1.emplace_back(first);
            status[abs(l[first].data)] = 1; // 标记此数的绝对值已经出现
        } else
            res2.emplace_back(first);
        first = l[first].next;
    }
    for (int i = 0; i != res1.size() - 1; ++i)
        cout << setw(5) << setfill('0') << res1[i] << " " << l[res1[i]].data << " " << setw(5) << setfill('0')
             << res1[i + 1] << "\n";
    cout << setw(5) << setfill('0') << res1[res1.size() - 1] << " " << l[res1[res1.size() - 1]].data << " " << -1
         << "\n";
    if (!res2.empty()) {
        for (int i = 0; i != res2.size() - 1; ++i)
            cout << setw(5) << setfill('0') << res2[i] << " " << l[res2[i]].data << " " << setw(5) << setfill('0')
                 << res2[i + 1] << "\n";
        cout << setw(5) << setfill('0') << res2[res2.size() - 1] << " " << l[res2[res2.size() - 1]].data << " " << -1;
    }
    return 0;
}