Question

我正在尝试创建一个将使用LinkedLists填充的LinkedList。主LinkedList将包含字符名称和描述，以及其内部LinkedList中保存的节点数。内部LinkedList将保存它们出现的章节编号以及该章节的简要说明。因此，例如，角色Joe（名字）是一个国王（描述符）并出现在第4章7和10章中。因此，他将有3个内部节点，描述他在这些章节中所做的事情。我不认为我正在添加到列表中，因为当我打电话来遍历列表并打印所有内容时，我只会得到我添加的第一个人。

我做的两个结构。

struct Person{
    char * name;
    char * descriptor;
    int count;
    struct Person * next;
    struct Information * info_head;
};
struct Information{
    char * text;
    int chapter;
    struct Information * next;
};

创建指针。

struct Person * new_person, *temp_pers, *head_pers;
struct Information * new_info, *temp_info, *head_info;

用于添加新字符的功能。

void add(char * name, char * descriptor, char * info, int chapter){
new_person = (struct Person *)malloc(sizeof(struct Person));
new_info = (struct Information *)malloc(sizeof(struct Information));

new_info->chapter = chapter;
new_info->text = info;
new_person->name = name;
new_person->descriptor = descriptor;

if(head_pers == NULL){  //List is empty
    head_pers = new_person; //add new person
    new_person->info_head = new_info;//link to information
    head_pers->next = NULL;
    head_info = new_info; //name that information start of information list
    head_pers->count = 1;
    head_info->next = NULL;
}
else{
    temp_pers = head_pers;
    temp_info = head_info;
    while(temp_pers != NULL){ //iterate through list of people
        if(strcmp(temp_pers->name, name) == 0){ //adding a duplicate
            while(temp_info != NULL){ //iterate through that persons info list
                temp_info = temp_info->next;
            } //reached the end of the list
            temp_info = new_info;
            temp_pers->count = temp_pers->count + 1;
            temp_pers->next = NULL;
        }
        temp_pers = temp_pers->next;
    }
    //reached end of persons list with no find
    //add new person to the end of the list
    temp_pers = new_person;
    temp_pers->count = temp_pers->count + 1;
    temp_pers->next = NULL;
}

}

我的测试打印方法

void printAll(){
    temp_pers = head_pers;
    temp_info = head_info;
    while(temp_pers != NULL){
        printf("%s, %s %d\n", temp_pers->name, temp_pers->descriptor, temp_pers->count);
        while(temp_info != NULL){
            printf("%d\t%s", temp_info->chapter, temp_info->text);
            temp_info = temp_info->next;
        }
        temp_pers = temp_pers->next;
    }
}

主要方法

int main(){
    add("Joe", "the King", "had a child.", 4);
    add("Joe", "the King", "started a war", 7);
    add("Sue", "the Queen", "poisoned Joe", 10);

    printAll();

    return 0;
}

处理LinkedLists的LinkedList令我感到困惑，也许我只是错过了一些非常小的东西，或者可能是非常大的东西，但任何帮助都会很棒。几乎忘了，代码确实编译并输出了这个......

    Joe, the King 2
4   had a child.

因为它将Joe的计数打印为2，我认为它正在解决这个问题。

Answer 1

除pers_head之外的所有全局变量都应该是本地变量。只有一个头，即人名单的头部。所有其他信息都包含在此列表中：其他人与next指针竞争;通过此人info的信息。最重要的是，没有全球信息主管;信息头属于每个人。

当您打印人员和事件列表时，您应该使用两个循环：

void printAll()
{
    struct Person *pers = head;

    while (pers) {
        printf("%s, %s [%d]\n",
            pers->name, pers->desc, pers->count);

        struct Information *info = pers->info;

        while (info) {
            printf("    %d: %s\n", info->chapter, info->text);
            info = info->next;
        }

        pers = pers->next;
    }
}

请注意pers和info是局部变量，而不是全局变量。它们被使用并且仅在它们被定义的范围内具有重要性。这很好：很容易看出pers只是迭代所有人，没有别的。

添加人员时，您可以立即创建新的人员节点。但如果此人已在列表中，您可能不需要新节点。仅在您确实需要时才创建节点。

你的add函数做了太多事情：它分配并填充节点和信息，并组织列表。如果您编写单独的函数来创建节点并向个人插入信息，核心列表代码将看起来更清晰。

可能的实现（稍微改变或缩短变量名称）可以是：

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

struct Person{
    char *name;
    char *desc;
    int count;
    struct Person *next;
    struct Information *info;
};
struct Information{
    char *text;
    int chapter;
    struct Information *next;
};

struct Person *head;

struct Information *info_new(char *text, int chapter)
{
    struct Information *info = malloc(sizeof(*info));

    info->text = text;
    info->chapter = chapter;
    info->next = NULL;

    return info;
}

struct Person *pers_new(char *name, char *desc)
{
    struct Person *pers = malloc(sizeof(*pers));

    pers->name = name;
    pers->desc = desc;
    pers->next = NULL;
    pers->info = NULL;
    pers->count = 0;

    return pers;
}

void pers_add_info(struct Person *pers, struct Information *info)
{
    if (pers->info == NULL) {
        pers->info = info;
    } else {
        struct Information *j = pers->info;

        while (j->next) j = j->next;
        j->next = info;
    }

    info->next = NULL;
    pers->count++;
}

void add(char *name, char *desc, char *infotext, int chapter)
{
    struct Information *info = info_new(infotext, chapter);
    struct Person *pers = head;
    struct Person *prev = NULL;

    while (pers) {
        if (strcmp(pers->name, name) == 0
         && strcmp(pers->desc, desc) == 0) {
            pers_add_info(pers, info);
            return;
        }
        prev = pers;
        pers = pers->next;
    }

    pers = pers_new(name, desc);
    pers_add_info(pers, info);

    if (prev) {
        prev->next = pers;
    } else {
        head = pers;
    }
}

void printAll()
{
    struct Person *pers = head;

    while (pers) {
        printf("%s, %s [%d]\n",
            pers->name, pers->desc, pers->count);

        struct Information *info = pers->info;

        while (info) {
            printf("    %d: %s\n", info->chapter, info->text);
            info = info->next;
        }

        pers = pers->next;
    }
}

int main()
{
    add("Joe", "the king", "had a child", 4);
    add("Sue", "the queen", "gave birth to a child", 4);
    add("Ben", "the prince", "is born", 4);
    add("Joe", "the king", "started a war", 7);
    add("Joe", "the king", "returns home victorious", 8);
    add("Ben", "the prince", "is squire to Lord Sam", 8);
    add("Sam", "High Lord", "takes Sam as apprentice", 8);
    add("Ben", "the prince", "goes on a quest", 9);
    add("Sue", "the queen", "poisoned Joe", 10);
    add("Ben", "the prince", "takes revenge", 10);
    add("Sam", "High Lord", "goes on a crusade", 11);
    add("Sue", "the queen", "repents", 14);
    add("Sue", "the hermit", "dies of old age and lonely", 14);

    printAll();

    return 0;
}

Answer 2

如您所见，您的代码打印出第一个节点和子节点：("Joe", "the King", "had a child.", 4)，这意味着您的插入工作正常，至少在开始时。在此行之后它终止，意味着temp_pers = temp_pers->next == NULL;

现在，让我们来看看你的第二次插入：

else{
temp_pers = head_pers;
temp_info = head_info;
while(temp_pers != NULL){ //iterate through list of people
    if(strcmp(temp_pers->name, name) == 0){ //adding a duplicate
        while(temp_info != NULL){ //iterate through that persons info list
            temp_info = temp_info->next;
        } //reached the end of the list
        temp_info = new_info;
        temp_pers->count = temp_pers->count + 1;
        temp_pers->next = NULL;
    }
    temp_pers = temp_pers->next;
}
//reached end of persons list with no find
//add new person to the end of the list
temp_pers = new_person;
temp_pers->count = temp_pers->count + 1;
temp_pers->next = NULL;
}

您创建了temp_pers并为其分配了一些内容，但在范围结束后，此节点未与您的head_pers相关联。因此，每次构建新结构并将其分配给temp_pers时，您都不会将其输入主链表（又名 head_pers），因此每次检查时你在循环中的条件（这一个：while(temp_pers != NULL)）你检查你创建的最后一个节点，因为它没有链接到某个东西，它会在下一个节点上给你NULL。

如何解决：其他部分中的head_pers->next = temp_pers;。现在，这将确保第二个节点连接到第一个节点。从现在开始，您创建的每个节点都应该连接到列表的末尾。你可以通过保存最后一个节点（O（1）时间），或者通过在每次添加（O（n）时间）上遍历列表来实现这一点

Answer 3

在add中，在循环后，temp_pers指向NULL。您需要保留指向列表中最后一个人的指针：

struct Person *last_pers;

last_pers = temp_pers;

while(temp_pers != NULL){ //iterate through list of people
    if(strcmp(temp_pers->name, name) == 0){ //adding a duplicate
        while(temp_info != NULL){ //iterate through that persons info list
            temp_info = temp_info->next;
        } //reached the end of the list
        temp_info = new_info;
        temp_pers->count = temp_pers->count + 1;
        temp_pers->next = NULL;
    }
    last_pers = temp_pers;
    temp_pers = temp_pers->next;
}
//reached end of persons list with no find
//add new person to the end of the list
last_pers->next = new_person;  
new_person->count = 1;
new_person->next = NULL;

LinkedLists的LinkedList C.

3 个答案: