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 100010
struct Node{
int add;
int data;
int next;
};
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif

vector<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_JUDGE
fclose(stdin);
#endif
return 0;
}