题目例如以下：
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 (<= 105) 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 33218Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
这是一道著名坑题，假设真的用单向链表来做，实在是太变态了，以下给出一种利用map和vector完毕的简单方法。
这道题最大的坑在于有无效结点，也就是说除了从头结点走到-1的所有结点外。还有其它子链表，排除的方法非常easy，仅仅要从头到尾走一遍记录下来，其它的所有扔掉。
要解决问题，须要考虑下面几个方面：
①怎样依据题目输入存储结点？
採用结构体数组。规模为10万，下标即为自己的地址。结构体内存储编号和下一个地址。
②怎样进行反转？
仅仅用vector记录有效结点的编号。同一时候在记录时用map记录编号到地址的相应关系。
仅仅反转vector中的编号。
③怎样连接地址输出？
通过反转后的vector。结合map查询地址。实现地址连接。
一定要注意处理K=N和N=1的情况，注意地址的前导0，-1不能有前导0。
代码例如以下：
#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
using namespace std;
struct Node{
int me;
int num;
int next;
}nodes[100000];
int main()
{
int N;
map<int,int> addMap;
int add,num,next;
int head,K;
cin >> head >> N >> K;
for(int i = 0; i < N; i++){
scanf("%d%d%d",&add,&num,&next);
nodes[add].me = add;
nodes[add].num = num;
nodes[add].next = next;
addMap[num] = add;
}
vector<int> validList;
add = head;
while(add != -1){
Node n = nodes[add];
validList.push_back(n.num);
add = n.next;
}
vector<int> reverseList;
int len = validList.size();
int cur = len;
for(int group = 0; group * K < len; group++){
cur = group * K;
if(len - cur < K) break;
else{
cur = len;
for(int i = K - 1; i >= 0; i--){
reverseList.push_back(validList[group * K + i]);
}
}
}
for(int i = cur; i < len; i++){
reverseList.push_back(validList[i]);
}
len = reverseList.size();
if(len == 1){
int num = reverseList[0];
printf("%05d %d -1
",addMap[num],num);
}else{
for(int i = 0; i < len - 1; i++){
int now = reverseList[i];
int next = reverseList[i+1];
printf("%05d %d %05d
",addMap[now],now,addMap[next]);
}
int now = reverseList[len - 1];
printf("%05d %d -1
",addMap[now],now);
}
return 0;
}