题目如下：
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
whereAddressis the position of the node,Datais an integer, andNextis 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218Sample Output:00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1
这是一道著名坑题，如果真的用单向链表来做，实在是太变态了，下面给出一种利用map和vector完成的简单方法。
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，还有其他子链表，排除的方法很简单，只要从头到尾走一遍记录下来，其他的全部扔掉。
要解决这个问题，，需要考虑以下几个方面：
①如何根据题目输入存储结点？
采用结构体数组，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如何进行反转？
只用vector记录有效结点的编号，同时在记录时用map记录编号到地址的对应关系。
只反转vector中的编号。
③如何连接地址输出？
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。
代码如下：
#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\n",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\n",addMap[now],now,addMap[next]);}int now = reverseList[len – 1];printf("%05d %d -1\n",addMap[now],now);}return 0;}
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