1. WIKI教程
2. 通过实例学习C.
3. 将链接列表拆分为两个(Split Linked List into two)

将链接列表拆分为两个(Split Linked List into two)

实现 (Implementation)

该算法的实现如下 -

#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(list == NULL) {
      list = link;
      list->next = link;
      return;
   }
   current = list;
   while(current->next != list)
      current = current->next;
   // Insert link at the end of the list
   current->next = link; 
   link->next = list;
}
void display(struct node *head) {
   struct node *ptr = head;
   printf("[head] =>");
   //start from the beginning
   while(ptr->next != head) {        
      printf(" %d =>",ptr->data);
      ptr = ptr->next;
   }
   printf(" %d =>",ptr->data);
   printf(" [head]\n");
}
void split_list() {
   int count = 0;
   // Allocate memory for new node;
   struct node *list1;
   struct node *link;
   struct node *current;
   list1 = list;
   while(list1->next != list) {
      struct node *link = (struct node*) malloc(sizeof(struct node));
      link->data = list1->data;
      link->next = NULL;
      if(list1->data%2 == 0) {
         if(even == NULL) {
            even = link;
            even->next = link;
            list1 = list1->next;
            continue;
         } else {
            current = even;
            while(current->next != even) {
               current = current->next;
            }
            // Insert link at the end of the list
            current->next = link; 
            link->next = even;
         }
         list1 = list1->next;
      } else {
         if(odd == NULL) {
            odd = link;
            odd->next = link;
            list1 = list1->next;
            continue;
         } else {
            current = odd;
            while(current->next!= odd) {
               current = current->next;
            }
            // Insert link at the end of the list
            current->next = link; 
            link->next = odd;
         }
         list1 = list1->next;
      }
   }
   // Lets handle the last node
   link = (struct node*) malloc(sizeof(struct node));
   link->data = list1->data;
   link->next = NULL;
   if(list1->data%2 == 0) {
      current = even;
      while(current->next != even) {
         current = current->next;
      }
      // Insert link at the end of the list
      current->next = link; 
      link->next = even;
   } else {
      current = odd;
      while(current->next!= odd) {
         current = current->next;
      }
      // Insert link at the end of the list
      current->next = link; 
      link->next = odd;
   }
}
int main() {
   int i;
   for(i = 1; i <= 10; i++)
      insert(i);
   printf("Complete list: \n");
   display(list);
   split_list();
   printf("\nOdd  : ");
   display(odd);
   printf("Even : ");
   display(even);
   return 0;
}

输出 (Output)

该方案的产出应该是 -

Complete list:
[head] => 1 => 2 => 3 => 4 => 5 => 6 => 7 => 8 => 9 => 10 => [head]
Odd  : [head] => 1 => 3 => 5 => 7 => 9 => [head]
Even : [head] => 2 => 4 => 6 => 8 => 10 => [head]