PAT甲级——A1052 Linked List Sorting-白红宇

PAT甲级——A1052 Linked List Sorting

阅读量：4540 次

发布时间：2019-06-08

本文共 2513 字，大约阅读时间需要 8 分钟。

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

Then

Address Key Next

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

Output Specification:

For each test case, the output format is the same as that of the input, where

Sample Input:

5 0000111111 100 -100001 0 2222233333 100000 1111112345 -1 3333322222 1000 12345

Sample Output:

5 1234512345 -1 0000100001 0 1111111111 100 2222222222 1000 3333333333 100000 -1

注意点

（1）题目给出的结点中可能有不在链表中的无效结点

（2）最后一个测试点测试的是首地址为 - 1，即为空链表的情况，此时只输出0 - 1

（3）输出时结点地址除 - 1外要有5位数字，不够则在高位补0 。所以地址 - 1要进行特判输出

1 #include 
     
       2 #include 
       3 using namespace std; 4 int main() 5 { 6     int N, head, maxV = -1000000; 7     cin >> N >> head; 8     map
       
        value; 9     map
        
         >data;10     for (int i = 0; i < N; ++i)11     {12         int addr, val, next;13         cin >> addr >> val >> next;14         data.insert(make_pair(addr, make_pair(val, next)));15     }16     if (data.find(head) == data.end())17     {18         cout << 0 << " " << -1 << endl;19         return 0;//该链表为空20     }21     while (head != -1)//遍历一下数据，找到是链表中的数据22     {23         value.insert(make_pair(data[head].first, head));24         head = data[head].second;25     }26     printf("%d %05d\n", value.size(), value.begin()->second);27     for (auto ptr = value.begin(); ptr != value.end(); ++ptr)28     {29         if (ptr == value.begin())30             printf("%05d %d ", ptr->second, ptr->first);31         else32             printf("%05d\n%05d %d ", ptr->second, ptr->second, ptr->first);33     }34     printf("%d\n", -1);35     return 0;36 }

转载于:https://www.cnblogs.com/zzw1024/p/11281097.html

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