02-线性结构3 Reversing Linked List(25 分)
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 (≤10
5
) 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
java过不去
#include<iostream>
#include<cstring>
using namespace std;
struct data
{
int add;
int d;
int next;
};
int main()
{
int first,N,K,m=0,num=0;
cin>>first>>N>>K;
data a[100000];
data p[100000];
data aa[100000];
data res[100000];
int i,j;
for(i=0;i<N;i++)
{
cin>>a[i].add;
cin>>a[i].d;
cin>>a[i].next;
p[a[i].add].d=a[i].d;
p[a[i].add].next=a[i].next;
}
i=0;
while(first!=-1)
{
aa[i].add=first;
aa[i].d=p[first].d;
aa[i].next=p[first].next;
first=p[first].next;
num++;
i++;
}
N=num;//可能有节点不在链表上
/*赋值法for(i=0;i<N/K;i++)
{
for(j=K*i;j<K*(i+1);j++)
{
res[m]=aa[K*(2*i+1)-1-j];
m++;
}
}
for(i=m;i<N;i++)
{
res[i]=aa[i];
}*/
data temp;
for(i=0;i<N/K;i++)
{
for(j=K*i;j<K*i+K/2;j++)
{
temp=aa[K*(2*i+1)-1-j];
aa[K*(2*i+1)-1-j]=aa[j];
aa[j]=temp;
}
}
for(i=0;i<N-1;i++)
{
printf("%05d %d %05d\n",aa[i].add,aa[i].d,aa[i+1].add);
}
printf("%05d %d %d\n",aa[i].add,aa[i].d,-1);
return 0;
}
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