C語言連線合併兩個雙向連結串列程式

建立兩個雙向連結串列，並連線這兩雙向連結串列中的資料項連線合併成一個雙向連結的範例程式，將以下程式碼儲存到一個原始檔中：combine_two_doubly_linked_list.c， 如下所示 -

#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;
}

執行上面程式，得到以下結果 -

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]