PAT甲级真题 1052 Linked List Sorting (25分) C++实现（模拟链表）

题目

A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive N (< 105) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Key Next

where Address is the address of the node in memory, Key is an integer in [-105, 105], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.

Output Specification:

For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.

Sample Input:

5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345

Sample Output:

5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1

思路

用结构体存储节点信息，用容量为100000的vector模拟map存储地址到节点的映射关系(addr为小于100000的正整数，可用vector以空间换时间)。

先将所有节点信息保存到vector中。从head开始搜索链表，将相互关联的节点保存到新的vector中（里面可能有不在链表中的节点）。

对新的vector按data排序，再按规则顺序输出即可。这里注意输出技巧，将每一行的next和下一行的addr、data做为一组，最前端和最后端分别加上数组大小和-1即可。

代码

#include 
#include 
#include 
using namespace std;

struct Node{
     
    int addr;
    int data;
    int next;
};
bool cmp(Node &n1, Node &n2){
     
    return n1.data < n2.data;
}
int main(){
     
    int n, head;
    cin >> n >> head;
    vector<Node> mmap(100000);
    while (n--){
     
        Node node;
        cin >> node.addr >> node.data >> node.next;
        mmap[node.addr] = node;
    }
    vector<Node> nodes;
    while(head!=-1){
     
        nodes.push_back(mmap[head]);
        head = mmap[head].next;
    }
    sort(nodes.begin(), nodes.end(), cmp);
    printf("%d ", (int)nodes.size());
    for (int i=0; i<nodes.size(); i++){
     
        printf("%05d\n%05d %d ", nodes[i].addr, nodes[i].addr, nodes[i].data);
    }
    printf("-1\n");
    return 0;
}