02-线性结构3 Reversing Linked List (25 分)

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 (≤10​5​​) which is the total number of nodes, and a positive K (≤N) 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

Output:

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

解题：
1.对于5位输出结果00010，这种，解决的办法是采用如下代码，其中5代表5个长度，0代表着前面自动补0

printf("%05d",Address);

2.一开始我以为要用链表做，觉得在构造链表的时候太难，于是乎觉得采用数组，这道题也比较明显是结构体数组；
其实我一开始觉得链表就是List Node,申请节点什么的，但是后来发现其实链表是一种物理存储单元上非连续、非顺序的存储结构。采用数组也可以达到这种结果，我们只要不按顺序存储即可。
未完待续..