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
#include<iostream> #include<vector> #include<algorithm> #include<queue> #include<string> #include<map> #include<set> #include<stack> #include<string.h> #include<cstdio> #include <unordered_map> #include<cmath> using namespace std; struct Node { int address,data,next; }; int main() { int head,n,k; scanf("%d%d%d",&head,&n,&k); map<int,Node> mp; //Node temp; int a,b,c; for(int i=0;i<n;i++) { scanf("%d%d%d",&a,&b,&c); mp[a]={a,b,c}; } vector<Node>node; for(int i=head;i!=-1;i=mp[i].next) node.push_back(mp[i]); int len=node.size(); for(int i=0;i<len;i+=k) { if(len-i<k) break; int temp=1; for(int j=i;j<(i+i+k)/2;j++) { swap(node[j],node[i+k-temp]); temp++; } } for(int i=0;i<len-1;i++) { printf("%05d %d %05d ",node[i].address,node[i].data,node[i+1].address); } printf("%05d %d -1 ",node[len-1].address,node[len-1].data); return 0; }