02-1. Reversing Linked List (25)
http://www.patest.cn/contests/mooc-ds/02-1
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 33218Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
#define MAXN 100001
typedef struct{
int addr;
int data;
int next;
}Node;
Node nodes[MAXN];
vector<Node> list;
int main(){
int firstAdd, n, k;
scanf("%d%d%d", &firstAdd, &n, &k);
while(n--){
Node nn;
scanf("%d%d%d", &nn.addr, &nn.data, &nn.next);
nodes[nn.addr] = nn;
}
int address = firstAdd;
while(address != -1){
list.push_back(nodes[address]);
address = nodes[address].next;
}
int length = list.size();
int round = length/k;
for(int i = 1; i <= round; ++i){
int start = (i-1)*k;
int end = i*k;
reverse(list.begin() + start, list.begin() + end);
}
for(int i = 0; i < length-1; ++i){
printf("%05d %d %05d\n", list[i].addr, list[i].data, list[i+1].addr);
}
printf("%05d %d %d\n",list[length-1].addr, list[length-1].data, -1);
return 0;
}