02-线性结构3 Reversing Linked List (25 分)
Given a constant and a singly linked list , you are supposed to reverse the links of every elements on . For example, given being 1→2→3→4→5→6, if =3, then you must output 3→2→1→6→5→4; if =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 which is the total number of nodes, and a positive 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 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
#include <stdlib.h>
#include <stdio.h>
typedef struct node {
int Address;
int Data;
int Next;
}list;
void Array(list a[], int N, int headaddress);
void Exchange(list a[], int i, int j);
void Reverse(list a[], int N, int K);
int main(){
int K, N, headaddress, eff;
scanf("%d %d %d", &headaddress, &N, &K);
list a[N];
for (int i = 0; i<N; i++) {
scanf("%d %d %d", &a[i].Address, &a[i].Data, &a[i].Next);
}
Array(a, N, headaddress);
for (int i = 0; i<N; i++) {
if (a[i].Next == -1) {
eff = i + 1;
}
}
Reverse(a, eff, K);
for (int i = 0; i<eff; i++) {
a[i].Next = a[i + 1].Address;
if (i == eff - 1) {
a[i].Next = -1;
}
}
for (int i = 0; i<eff - 1; i++) {
printf("%05d %d %05d\n", a[i].Address, a[i].Data, a[i].Next);
}
printf("%05d %d %d\n", a[eff - 1].Address, a[eff - 1].Data, -1);
return 0;
}
void Array(list a[], int N, int headaddress) {
for (int i = 0; i<N; i++) {
if (a[i].Address == headaddress) {
Exchange(a, i, 0);
break;
}
}
for (int i = 0; i<N; i++) {
for (int j = i + 1; j<N; j++) {
if (a[i].Next == a[j].Address) {
Exchange(a, i + 1, j);
break;
}
}
}
}
void Exchange(list a[], int i, int j) {
list temp;
temp.Address = a[j].Address;
temp.Data = a[j].Data;
temp.Next = a[j].Next;
a[j].Address = a[i].Address;
a[j].Data = a[i].Data;
a[j].Next = a[i].Next;
a[i].Address = temp.Address;
a[i].Data = temp.Data;
a[i].Next = temp.Next;
}
void Reverse(list a[], int N, int K) {
int cnt = N / K;
if (cnt) {
for (int j = 1; j <= cnt; j++) {
for (int i = 1; i <= K; i++) {
if ((j - 1)*K + i - 1 <= j * K - i - 1) {
Exchange(a, (j - 1)*K + i - 1, j*K - i);
}
}
}
}
}