# 02-线性结构3 Reversing Linked List (PTA)

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

Python 的代码在第五个测试点“ 最大N,最后剩K-1不反转 ”，运行超时。C++代码可以AC。

d={}

start_addr, L, K = input().split(' ')
L = int(L)
K = int(K)

for i in range(L):

count = 0
for i in range(L):
break
else:
count += 1

res_index = 0
while(res_index + K <= count):
res_index += K

for i in range(len(addr_list) - 1):
print(addr_list[-1],d[ addr_list[-1] ][0],'-1') 

#include <iostream>
#include <algorithm>
#define MAXSIZE 100000
using namespace std;

struct LNode
{
int data;
int next;
}node[MAXSIZE];

int List[MAXSIZE];

int main()
{

for(int i = 0; i < n; i++)
{
cin >> data;
cin >> next;
}

int count = 0;

while(p_next!=-1)
{
List[count] = p_next;
p_next = node[p_next].next;
count = count + 1;
}

int first = 0;
while (first + k <= count)
{
reverse(&List[first], &List[first + k]);
first += k;
}

for(first = 0; first < count-1; first++)
printf("%05d %d %05d\n", List[first], node[List[first]].data, List[first + 1]);
printf("%05d %d -1\n", List[first], node[List[first]].data);
return 0;
}