PAT甲级——A1074 Reversing Linked List

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

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 (≤) which is the length of the sublist to be reversed. 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, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

算法设计：
由于所给的地址是5位非负整数，可以定义两个维度为100005的一维数组data、Next，负责储存数据域和下一个地址
定义一个vector<int>listAddress，由所给链表开始地址处开始遍历整个链表，按遍历顺序将各结点地址储存到listAddress中。
按要求对listAddress数组进行翻转，可以利用c语言库里的reverse函数

按格式要求进行结果输出

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）翻转时要注意如果最后一组要翻转的结点数量小于K，则不进行翻转；如果等于K，需要进行翻转

（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

//s首先说明一下PAT的常见陷阱，就是给出的数据未必是一条链表，可能是多条，
//所以首先从输入的数据的中找到那条链表
//巨简单，使用vector反转就行了
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int head, N, K;
int nodes[100100], nexts[100100];
vector<int>list;
int main()
{
    cin >> head >> N >> K;
    int adrr, data, next, temp = head;
    for (int i = 0; i < N; ++i)
    {
        cin >> adrr >> data >> next;
        nodes[adrr] = data;
        nexts[adrr] = next;
    }
    while (temp != -1)//找出这条链表
    {
        list.push_back(temp);
        temp = nexts[temp];
    }
    for (int i = K; i <= list.size(); i += K)
        reverse(list.begin() + i - K, list.begin() + i);
    for (int i = 0; i < list.size(); ++i)
    {
        printf("%05d %d ", list[i], nodes[list[i]]);
        if (i < list.size() - 1)
            printf("%05d
", list[i+1]);
        else
            printf("-1
");
    }
    return 0;
}