1074. Reversing Linked List (25)

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

IDEA

1.注意：可能给出的节点有不在链表上的情况，则这样的几点是不输出的，也就不能用n/k来确定进行了几次翻转

2.思路：将给出的几点按顺序牌号，然后按要求翻转，最后需要把最后一个节点的next赋值为-1

CODE

#include<iostream>#include<cstdio>#include<vector> #include<queue>#include<algorithm>#include<fstream>using namespace std;#define Max 100010struct Node{int add;int data;int next;};int main(){#ifndef ONLINE_JUDGEfreopen("input.txt","r",stdin);#endifvector<Node> vec(Max);int start,n,k;cin>>start>>n>>k;for(int i=0;i<n;i++){Node node;cin>>node.add>>node.data>>node.next;vec[node.add]=node;}vector<Node> tmp;int p=start;while(p!=-1){tmp.push_back(vec[p]);p=vec[p].next;}//这样写最后一个测试点通不过，没有考虑有的节点可能不在链表上的情况，不能用n/k确定翻转几次 //int m=n/k;//for(int i=0;i<m;i++){//reverse(tmp.begin()+i*k,tmp.begin()+(i+1)*k);//}for(int i=0;i+k<=tmp.size();i+=k){reverse(tmp.begin()+i,tmp.begin()+i+k);}for(int i=0;i<tmp.size()-1;i++){tmp[i].next=tmp[i+1].add;}//tmp[tmp.size()-1].next=-1;for(int i=0;i<tmp.size();i++){if(i!=tmp.size()-1){printf("%05d %d %05d\n",tmp[i].add,tmp[i].data,tmp[i].next);}else{printf("%05d %d -1\n",tmp[i].add,tmp[i].data);}}#ifndef ONLINE_JUDGEfclose(stdin);#endifreturn 0;}