1133 Splitting A Linked List

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤) which is the total number of nodes, and a positive K (≤). The address of a node is a 5-digit nonnegative 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 an integer in [, and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

 

Sample Output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

题意：

　　给出一个自定义的链表，将链表按照要求的顺序输出。

思路：

　　构造一个结构体来表示节点，开一个数组用下标来表示结点的地址，根据结点地址遍历数组，将符合要求的结点依次放入ans中，最后将ans中的数据输出即可。

Code：

#include <bits/stdc++.h>

using namespace std;

struct Node {
    int address;
    int date;
    int next;
};

int main() {
    int first, n, k;
    cin >> first >> n >> k;
    vector<Node> v(100005), ans;
    int address, date, next;
    for (int i = 0; i < n; ++i) {
        cin >> address >> date >> next;
        v[address] = {address, date, next};
    }
    address = first;
    for (; address != -1; address = v[address].next)
        if (v[address].date < 0) ans.push_back(v[address]);

    address = first;
    for (; address != -1; address = v[address].next)
        if (v[address].date >= 0 && v[address].date <= k)
            ans.push_back(v[address]);
    address = first;
    for (; address != -1; address = v[address].next)
        if (v[address].date > k) ans.push_back(v[address]);
    printf("%05d %d", ans[0].address, ans[0].date);
    for (int i = 1; i < ans.size(); ++i) {
        printf(" %05d
%05d %d", ans[i].address, ans[i].address, ans[i].date);
    }
    printf(" -1
");
    return 0;
}