Reversing Linked line
Reversing Linked List
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.
小白白先写一小段代码:
#include<iostream> using namespace std; struct node { int data; node* next; }; struct stack { int* array; int size; int top; }; void creat_stack(stack*& x, int n) { x->array = new int[n]; memset(x->array, 0, sizeof(int) * n); x->size = n; x->top = -1; } bool Is_empty(stack*& x) { return (x->top == -1); } bool Is_full(stack*& x) { return (x->top == x->size - 1); } void push(stack*& x, int a) { if (Is_full(x)) { cout << "The stack is full.\n"; return; } else { x->array[++x->top] = a; } } int pop(stack*& x) { if (Is_empty(x)) { cout << "The stack is empty.\n"; exit(0); } else { return x->array[x->top--]; } } void creat_node(node*& head, int n) { node* s = new node; node* p = new node; cin >> s->data; for (int i = 0; i < n; i++) { if (!head) { head = s; p = s; } else { s = new node; cin >> s->data; p->next = s; p = p->next; } } p->next = NULL; } void reverse(node*& head, int n, int k) { stack* s = new stack; creat_stack(s, k); node* p = head; node* q = head; int count = 0; while (k * ++count <= n && p && q) { for (int i = 0; i < k; i++) { push(s, p->data); p = p->next; } for (int i = 0; i < k; i++) { q->data = pop(s); q = q->next; } } } void read_node(node* head) { while (head) { cout << head->data; head = head->next; if (head)cout << "->"; } cout << endl; } int main() { int n = 0, k = 0; cin >> n >> k; node* head = NULL; creat_node(head,n); reverse(head, n, k); read_node(head); }
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
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被卡住了嘤嘤嘤,有空再来补坑
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
struct node
{
char add[10];
int data;
char nextadd[10];
node* next;
};
struct stack
{
node* array;
int size;
int top;
};
void creat_stack(stack*& x, int n)
{
x->array = new node[n];
memset(x->array, 0, sizeof(node) * n);
x->size = n;
x->top = -1;
}
bool Is_empty(stack*& x)
{
return (x->top == -1);
}
bool Is_full(stack*& x)
{
return (x->top == x->size - 1);
}
void push(stack*& x, node*& a)
{
if (Is_full(x))
{
cout << "The stack is full.\n";
return;
}
else
{
strcpy_s(x->array[++x->top].add, 10, a->add);
x->array[x->top].data = a->data;
strcpy_s(x->array[x->top].nextadd, 10, a->nextadd);
}
}
void pop(stack*& x, node*& a)
{
if (Is_empty(x))
{
cout << "The stack is empty.\n";
exit(0);
}
else
{
strcpy_s(a->add,10,x->array[x->top].add);
a->data = x->array[x->top].data;
strcpy_s(a->nextadd, 10, x->array[x->top--].nextadd);
}
}
void creat_node(node*& head, int n)
{
node* s = new node;
node* p = NULL;
cin >> s->add >> s->data >> s->nextadd;
for (int i = 0; i < n; i++)
{
if (!head)
{
head = s;
p = s;
}
else
{
s = new node;
cin >> s->add >> s->data >> s->nextadd;
p->next = s;
p = p->next;
}
}
p->next = NULL;
}
void swap_node(node*& p, node*& q)
{
node* temp = new node;
strcpy_s(temp->add, 10, q->add);
temp->data = q->data;
strcpy_s(temp->nextadd, 10, q->nextadd);
strcpy_s(q->add, 10, p->add);
q->data = p->data;
strcpy_s(q->nextadd, 10, p->nextadd);
strcpy_s(p->add, 10, temp->add);
p->data = temp->data;
strcpy_s(p->nextadd, 10, temp->nextadd);
}
void sort_node(node*& head, char* add)
{
node* p = head;
node* q = head;
while (strcmp(add, "-1"))
{
while (strcmp(p->add , add))
{
p = p->next;
}
strcpy_s(add,10,p->nextadd);
swap_node(p, q);
q = q->next;
p = head;
}
}
void reverse_node(node*& head, int n, int k)
{
stack* s = new stack;
creat_stack(s, k);
node* p = head;
node* q = head;
node* x = new node;
int count = 0;
while (k * ++count <= n )
{
for (int i = 0; i < k; i++)
{
push(s,p);
p = p->next;
}
pop(s, x);
strcpy_s(q->nextadd, 10, x->add);
if (q != head)q = q->next;;
for (int i = 1; i < k; i++)
{
strcpy_s(q->add, 10, x->add);
q->data = x->data;
pop(s, x);
strcpy_s(q->nextadd, 10, x->add);
q = q->next;
}
strcpy_s(q->add, 10, x->add);
q->data = x->data;
}
}
void read_node(node* head)
{
while (head)
{
cout << head->add << ' ' << head->data << ' ' << head->nextadd << endl;
head = head->next;
}
}
int main()
{
char* head_add = new char[10];
int n = 0, k = 0;
cin >> head_add >> n >> k;
node* head = NULL;
creat_node(head, n);
sort_node(head, head_add);
reverse_node(head, n, k);
read_node(head);
}
补坑了补坑了,好像写得有点儿乱哎,不知道有没有错。

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