組合兩個連結串列的程式

實現

此演算法的實現如下所示 −

#include <stdio.h>
#include <stdlib.h>

struct node {
   int data;
   struct node *next;
};

struct node *even = NULL;
struct node *odd = NULL;
struct node *list = NULL;

//Create Linked List
void insert(int data) {
   // Allocate memory for new node;
   struct node *link = (struct node*) malloc(sizeof(struct node));
   struct node *current;

   link->data = data;
   link->next = NULL;

   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; 
      }
   
   } 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; 
      }
   }
}

void display(struct node *head) {
   struct node *ptr = head;

   printf("[head] =>");
   
   while(ptr != NULL) {        
      printf(" %d =>",ptr->data);
      ptr = ptr->next;
   }

   printf(" [null]\n");
}

void combine() {
   struct node *link;

   list = even;
   link = list;
    
   while(link->next!= NULL) {
      link = link->next;
   }

   link->next = odd;
}

int main() {
   int i;

   for(i = 1; i <= 10; i++)
      insert(i);

   printf("Even : ");
   display(even);

   printf("Odd  : ");
   display(odd);

   combine();
   
   printf("Combined List :\n");
   display(list);
   
   return 0;
}

輸出

程式的輸出應為 −

Even : [head] => 2 => 4 => 6 => 8 => 10 => [null]
Odd  : [head] => 1 => 3 => 5 => 7 => 9 => [null]
Combined List :
[head] => 2 => 4 => 6 => 8 => 10 => 1 => 3 => 5 => 7 => 9 => [null]