PAT 甲级 1074 Reversing Linked List

1074 Reversing Linked List （25 point(s)）

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

Sample Output:

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

经验总结：

这一题，先遍历一遍链表，将不再此链表上的结点排除，然后，将其按照链表顺序排序，并且计算出链表结点数目，之后，以K为间距进行逆置（最后不满K个元素的尾部无需逆置），然后按顺序输出即可（注意链尾的输出），这种方法虽然速度不是很快，但是容易理解。

AC代码

#include 
#include 
using namespace std;
const int INF=0x3fffffff;
const int maxn=100010;
struct node
{
	int data,address,next,level;
	node():level(INF){}
}Node[maxn];
bool cmp(node a,node b)
{
	return a.level