02-线性结构2 Reversing Linked List

Given a constant KK and a singly linked list LL, you are supposed to reverse the links of every KK elements on LL. For example, given LL being 1→2→3→4→5→6, if K = 3K=3, then you must output 3→2→1→6→5→4; if K = 4K=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 NN (\le 10^5≤10

​5

​​ ) which is the total number of nodes, and a positive KK (\le N≤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 NN 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

思路：建立双向链表，依次访问直到数目为倒序的次数，若能访问，则用双向链表倒序输出，否则就正序输出。

#include <iostream>
#include <vector>
using namespace std;

typedef struct link
{
int rank;
int base_add;
int next_add;
link* parent;
link* child;
}link;

int main()
{
int begin_add, N, listnum;

cin >> begin_add;
cin >> N;
cin >> listnum;

vector<pair<int, int>> address(N);//first is base address,second is next address
int base_add, next_add, rank;
for (int i = 0; i < N; i++)
{
cin >> base_add;
cin >> rank;
cin >> next_add;

address[rank - 1].first = base_add;
address[rank - 1].second = next_add;
}

link* top;
link* current;
link* child;
top = (link*)malloc(sizeof(link));
current = top;
top->parent = NULL;
for (int i = 0; i < N; i++)
{
current->base_add = address[i].first;
current->next_add = address[i].second;
current->rank = i+1;
if (i != N - 1)//not the last node
{
child = (link*)malloc(sizeof(link));
current->child = child;
child->parent = current;
current = child;
}
else
{
current->child = NULL;
}
}
int count;
current = top;
bool flag = false;
while (current != NULL)
{
count = 1;
child = current;
if (flag)
printf("%05d\n", child->base_add);
while (current != NULL&&count!=listnum)
{
current = current->child;
count++;
}
if (count == listnum)
{
child = current;

for (int j = 0; j < listnum-1; j++)
{
//if (child->parent != NULL)
printf("%05d %d %05d\n", child->base_add, child->rank, child->parent->base_add);
child = child->parent;
}
printf("%05d %d ", child->base_add, child->rank);
current = current->child;
flag = true;
}
else
{
for (int j = 0; j < count-2; j++)
{
printf("%05d %d %05d\n", child->base_add, child->rank, child->child->base_add);
child = child->child;
}
printf("%05d %d %d\n", child->base_add, child->rank, -1);
break;
}

}

return 0;
}