C中的链接列表

Question

尽管这不是我课程的要求，但我正在尝试实现单链表和双链表，以确保我理解这些概念。 我已经制定了大部分我认为除了那我试图调试和一个问题printf我的出路。 尽管我阅读了不同的文章，但我无法找出它们为什么不能以类似的方式工作。

一个我的插入函数，当while(h != NULL)踢出时我必须再次检查最后一项，因为它会跳过它。 我想到了一个do / while循环，但是那行不通。

我对搜索功能有完全相同的循环，它在使用while(h != NULL)时找到列表中的最后一项， while(h != NULL)不是在处理它之前踢我。

这是从搜索功能开始的循环，

    while(h != NULL) //check list
        if(h->val == s)
            return true;
        else
           h = h->next; 

这是不是来自insert函数的循环，这个循环在退出循环之前不会最终使用链表中的最终值。 如果您在阅读完整代码之前必须检查是否if (t->next == NULL && t->val > n)来搜索最后一项。

        while(t->next != NULL) //assigns it in the middle
        {
            if(t->val < n)
                t = t->next;
            else
            {
                node* tt = t->prev; 
                tt->next = a;
                a->next = t;
                a->prev = t->prev;
                t->prev = a;

                return h; //returns original
            }
        }

在此先感谢，希望我能解释它。 当我尝试学习我不知道的东西时很难。

#include <stdio.h>
#include <stdlib.h> //for rand();
#include <stdbool.h> //bool function

typedef struct _node
{
    int val;
    struct _node* next;
    struct _node* prev;
} node;

node* create(int n, node* h);

node* insert(int n, node* h);

bool search(int s, node* h);

void print(node* h);

void del(node* h);

int main()
{
    node* h = NULL;

    for(int i = 0; i < 15; i++)
    {
        int n = rand() % 15;
        h = create(n, h);
    }

    print(h);

    int s = 10; //number to search for

    if(search(s, h))
        printf("Found Item\n");
    else
        printf("Item not found\n");

    del(h);
}

bool search(int s, node* h)
{
    if(h == NULL)
        printf("No nodes created to search.\n");

    while(h != NULL) //check list
        if(h->val == s)
            return true;
        else
           h = h->next; 

    return false;
}


void del(node* h)
{
    node* t = h;

    while (h->next != NULL)
    {
        t = h->next;
        h->next = NULL;
        free(h);
        h = t;
    }
        free(h);
}

node* insert(int n, node* h)
{
    node* a = malloc(sizeof(node));
    node* t = h;

    if(a == NULL)
    {
        printf("Failed to create new node\n");
        return NULL;
    }

    a->val = n; //sets new node to have value of n

    if(h->val >= n) //assigns the integer to the beginning
    {
        a->next = t;
        t->prev = a;
        a->prev = NULL;
        h = a; //assigns a new head since it appended the item to the beginning
        return h;
    }

    while(t->next != NULL) //assigns it in the middle
    {
        if(t->val < n)
            t = t->next;
        else
        {
            node* tt = t->prev; 
            tt->next = a;
            a->next = t;
            a->prev = t->prev;
            t->prev = a;

            return h; //returns original
        }
    }

    if (t->next == NULL && t->val > n) //check to see if second to last  
    {       
        node* tt = t->prev;
        tt->next = a;
        a->next = t;
        a->prev = t->prev;
        t->prev = a;
    }
    else
    {
        a->next = NULL;
        a->prev = t;
        t->next = a;
    }

    return h;
}

node* create(int n, node* h)
{
    if(h == NULL) //create first node in the list
    {
        h = malloc(sizeof(node));
        if(h == NULL)
        {
            printf("Failed to create first node\n");
            return NULL;
        }
        h->val = n;
        h->next = NULL;
        h->prev = NULL;
    }
    else //create additional nodes in the list
        h = insert(n, h);

    return h;
}

void print(node* h)
{
    if(h == NULL)
        printf("List is empty\n");
    else
        while(h != NULL)
        {
            printf("%i\n", h->val);
            h = h->next;
        }
}

Answer 1

如果我正确理解您的问题，那么您正在询问以下几行：

while(t->next != NULL) //assigns it in the middle
{
    if(t->val < n)
        t = t->next;
    else
    {
        node* tt = t->prev; 
        tt->next = a;
        a->next = t;
        a->prev = t->prev;
        t->prev = a;

        return h; //returns original
    }
}

if (t->next == NULL && t->val > n) //check to see if second to last  
{       
    node* tt = t->prev;
    tt->next = a;
    a->next = t;
    a->prev = t->prev;
    t->prev = a;
}
else
{
    a->next = NULL;
    a->prev = t;
    t->next = a;
}

问题是：“如何避免在片刻之后从片刻内部重复代码块”

也许是这样的：

while(1) 
{
    if(t->val < n)
    {
        if (t->next == NULL) break;
        t = t->next;
    } 
    else
    {
        //assigns it in the middle
        node* tt = t->prev; 
        tt->next = a;
        a->next = t;
        a->prev = t->prev;
        t->prev = a;

        return h; //returns original
    }
}

// Add new node at the end
a->next = NULL;
a->prev = t;
t->next = a;