pat-basic-1025-Reverse Linked List

Given a constant k and a single-linked table L, write a program that reverses every K-node in L. For example: Given L is 1→2→3→4→5→6,k 3, the output should be 3→2→1→6→5→4, and if K is 4, the output should be 4→3→2→1→5→6, that is, the last K elements do not invert.

Input format:

Each input consists of 1 test cases. The 1th row of each test case gives the address of the 1th node, the total number of nodes with positive integer N (<= 105), and the positive integer k (<=n), which is the number of sub-chain nodes that require inversion. The address of the node is a 5-bit nonnegative integer, and the null address is represented by-1.

Next there are n rows, each in the following format:

Address Data Next

Where address is a node location, data is the integer that the node holds, andnext is the address of the next node.

Output format:

For each test case, sequentially output the inverted list with one row on each node, with the same format as the input.

Input Sample:

00100 6 400000 4 9999900100 1 1230968237 6-133218 3 0000099999 5 6823712309 2 33218

Sample output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6-1


A reversal of every k element

/*
DATA:
1 6 3
1 1 2
2 2 3
3 3 4
4 4 5
5 5 6
6 6-1
*/

#include <bits/stdc++.h>#defineMAXN 100000+50using namespacestd; structdata{intaddress; intdata; intNext;  };  DATA ARR[MAXN]; Vector<DATA>ans; Stack<DATA>s; intsrcaddress, N, K; intMain () {scanf ("%d%d%d", &srcaddress, &n, &k);  for(inti =0; I < n; ++i) {          intPOS; scanf ("%d", &POS); Arr[pos].address=POS; scanf ("%d%d", &arr[pos].data, &arr[pos].next); }       Do{ans.push_back (arr[srcaddress]); Srcaddress=Arr[srcaddress].next; } while(Srcaddress! =-1); //printf ("%d\n", Ans.size ());     intSZ =ans.size (); //k = min (k, SZ);     intpos =0; /*for (int i = 0; i < K/2; ++i) {DATA tmp = ans[i];         Ans[i] = ans[(k-1)-i];     ans[(k-1)-i] = tmp; }*/       while(Pos+k <=SZ) {           while(!S.empty ())          {S.pop (); }           for(inti = pos; i < K+pos; ++i)          {S.push (ans[i]); }           while(!S.empty ()) {Ans[pos++] =S.top ();          S.pop (); }         //pos + = k;     }       for(inti =0; I < sz-1; ++i) {printf ("%05d%d%05d\n", Ans[i].address, Ans[i].data, ans[i+1].address); } printf ("%05d%d%d\n", ans[sz-1].address, ans[sz-1].data,-1); return 0; }  

Capouis ' CODE

pat-basic-1025-Reverse Linked List