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。

思路相同， 先读取有效的数据，排除不在链表上的节点 ，然后用有序列表的部分反转操作达到反转。Python效率不如C++，C++用4ms算完的，Python要22ms，最大长度C++花了200多ms，而这题限制400ms。也想不出优化的地方了。

编译器：Python(python3)

d={}

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

for i in range(L):
    addr1, data, next_addr = input().split(' ')
    d[addr1] = [data,next_addr]

addr_list = []
count = 0
temp_addr = start_addr 
for i in range(L):
    if temp_addr == '-1':
        break
    else:
        count += 1
        addr_list.append(temp_addr)
        temp_addr = d[temp_addr][1]

res_index = 0
while(res_index + K <= count):
    sub_list = addr_list[res_index:res_index+K]
    addr_list[res_index:res_index+K] = sub_list[::-1]
    res_index += K

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

编译器：C++(g++)

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

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

int List[MAXSIZE];

int main()
{
    int head_addr, n, k;
    cin >> head_addr >> n >> k;

    int addr, data, next;
    for(int i = 0; i < n; i++)
    {
        cin >> addr;
        cin >> data;
        cin >> next;
        node[addr].data = data;
        node[addr].next = next;
    }

    int count = 0;
    int p_next = head_addr;

    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;
}