结合两个双重链接列表（Combine Two Doubly Linked List）

实现 (Implementation)

该算法的实现如下 -

#include <stdio.h>
#include <stdlib.h>
struct node {
   int data;
   struct node *prev;
   struct node *next;
};
struct node *list = NULL;
struct node *list_last = NULL;
struct node *even = NULL;
struct node *even_last = NULL;
struct node *odd = NULL;
struct node *odd_last = NULL;
struct node *current = NULL;
//Create Linked List
void insert(int data) {
   // Allocate memory for new node;
   struct node *link = (struct node*) malloc(sizeof(struct node));
   link->data = data;
   link->prev = NULL;
   link->next = NULL;
   // If head is empty, create new list
   if(data%2 == 0) {
      if(even == NULL) {
         even = link;
         return;
      } else {
         current = even;
         while(current->next != NULL)
            current = current->next;
         // Insert link at the end of the list
         current->next = link; 
         even_last = link;
         link->prev = current;
      }
   } else {
      if(odd == NULL) {
         odd = link;
         return;
      } else {
         current = odd;
         while(current->next!=NULL)
            current = current->next;
         // Insert link at the end of the list
         current->next = link; 
         odd_last = link;
         link->prev = current;
      }
   }
}
//display the list
void print_backward(struct node *head) {
   struct node *ptr = head;
   printf("\n[last] <=>");
   //start from the beginning
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->prev;
   }
   printf(" [head]\n");
}
//display the list
void printList(struct node *head) {
   struct node *ptr = head;
   printf("\n[head] <=>");
   //start from the beginning
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->next;
   }
   printf(" [last]\n");
}
void combine() {
   struct node *link;
   list = even;
   link = list;
   while(link->next!= NULL) {
      link = link->next;
   }
   link->next = odd;
   odd->prev = link;
   // assign link_last to last node of new list
   while(link->next!= NULL) {
      link = link->next;
   }
   list_last = link;  
}
int main() {
   int i;
   for(i = 1; i <= 10; i++)
      insert(i);
   printf("Even : ");
   printList(even);
   printf("Even (R): ");
   print_backward(even_last);
   printf("Odd  : ");
   printList(odd);
   printf("Odd (R) : ");
   print_backward(odd_last);
   combine();
   printf("Combined List :\n");
   printList(list);
   printf("Combined List (R):\n");
   print_backward(list_last);
   return 0;
}

输出 (Output)

该方案的产出应该是 -

Even :
[head] <=> 2 <=> 4 <=> 6 <=> 8 <=> 10 <=> [last]
Even (R):
[last] <=> 10 <=> 8 <=> 6 <=> 4 <=> 2 <=> [head]
Odd  :
[head] <=> 1 <=> 3 <=> 5 <=> 7 <=> 9 <=> [last]
Odd (R) :
[last] <=> 9 <=> 7 <=> 5 <=> 3 <=> 1 <=> [head]
Combined List :
[head] <=> 2 <=> 4 <=> 6 <=> 8 <=> 10 <=> 1 <=> 3 <=> 5 <=> 7 <=> 9 <=> [last]
Combined List (R):
[last] <=> 9 <=> 7 <=> 5 <=> 3 <=> 1 <=> 10 <=> 8 <=> 6 <=> 4 <=> 2 <=> [head]