PAT 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
简单的模拟题,以空间换时间:
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std; const int NUM=; struct Node
{
int address;
int data;
int next;
}; Node nodes[NUM];
vector<Node> list;
int main()
{
int fnAddress,N,K;
scanf("%d %d %d",&fnAddress,&N,&K);
int i;
for(i=;i<N;++i)
{
Node node;
scanf("%d %d %d",&node.address,&node.data,&node.next);
nodes[node.address]=node;
}
int address=fnAddress;
while(address!=-)//去噪
{
list.push_back(nodes[address]);
address=nodes[address].next;
}
int size=list.size();
int round=size/K;
int start,end;
for(i=;i<=round;++i)
{
start=(i-)*K;
end=i*K;
reverse(list.begin()+start,list.begin()+end);
}
for(i=;i<size-;++i)
{
printf("%.5d %d %.5d\n",list[i].address,list[i].data,list[i+].address);
}
printf("%.5d %d %d\n",list[i].address,list[i].data,-);
return ;
}
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