文章标题

1074. Reversing Linked List
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

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

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下面开始分析题目：
1.题目意思是将链表每k个进行反转，不足k个的不反转，最后讲反转的链表输出。
难点如下：
2.首先判链表是否存在，如果输入的首地址等于-1，那么链表肯定不存在，程序直接return 0即可。
3.如果k值大于真实链表的大小，那么不需要进行反转，直接输出即可。
4.如果k值小于或者等于真实链表的大小，那么对反转的次数和每次反转的开始结束下标进行控制即可。
技术实现
5.链表的建立，我是用了哈希+vector，避免多次循环。用空间换取时间。
6.链表的输入，我采取了偷懒政策，我再交换节点的时候并没有真得对节点的next进行重新赋值，而是直接在输出的时候输出下一个节点的头地址。

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代码如下，代码中有备注，还算清晰

#include <iostream>#include <stdio.h>#include <vector>/* run this program using the console pauser or add your own getch, system("pause") or input loop */using namespace std;struct node{//节点结构体     int add;    int value;    int next;};int main(int argc, char *argv[]) {    int first,n,k;    scanf("%d%d%d",&first,&n,&k);    if(first == -1){//链表为空，什么也不输出         return 0;    }    node list[100010];//用于哈希，为了建表     for(int i=0;i<n;i++){        int add,value,next;        scanf("%d%d%d",&add,&value,&next);        list[add].add = add;        list[add].value = value;        list[add].next=next;    }    vector<node> reallist;//存储真实的链表     while(first!=-1){        int next = list[first].next;        reallist.push_back(list[first]);        first = next;    }//  printf("%d",reallist.size());    int realsize = reallist.size();    vector<node> result;//结果链表         int start = 0;        int end = 0;    if(realsize<k){//k值比真实的值大，因此不需要反转         for(int i=0;i<realsize-1;i++){            printf("%05d %d %05d\n",reallist[i].add,reallist[i].value,reallist[i+1].add);//输出的技巧         }        printf("%05d %d %d\n",reallist[realsize-1].add,reallist[realsize-1].value,-1);//输出最后一个     }    else{        int count = realsize/k;//可以反转的次数         start = 0;//第一次反转的开始位置         end = start + k-1;//第一次反转的结束位置         int index = 0;        for(int i=1;i<=count;i++){//可以反转的次数             for(int j=end;j>=start;j--){//每次反转的循环                 node t = reallist[j];                result.push_back(t);            }            start += k;//反转的开始             end = start+k-1;//反转的结束         }        for(int i = count*k;i<realsize;i++){//不能反转的             result.push_back(reallist[i]);        }    }    for(int i=0;i<result.size()-1;i++){//输出         printf("%05d %d %05d\n",result[i].add,result[i].value,result[i+1].add);    }    printf("%05d %d %d\n",result[realsize-1].add,result[realsize-1].value,-1);    return 0;}