《算法导论》第10章 基本数据结构 (2)链表


《算法导论》里实现的是无序双向链表

源文件 list.h:链表的接口,定义链表和结点的类型,以及链表支持的操作。

typedef int DATA_TYPE;

// 注意将结构声明为新类型的语法
struct tagNode {
     DATA_TYPE data;
     struct tagNode *prev, *next;
};
typedef struct tagNode Node;

typedef struct {
     Node *head;    
} List;


List * list_create(void);
Node * node_create(DATA_TYPE k);
void list_destory(List*);
void list_print(List*);

Node * list_search(List*, DATA_TYPE);
void list_insert(List*, Node*);
void list_delete(List*, Node*);




源文件 list.c:链表的实现。

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

List * list_create(void)
{    
     return malloc(sizeof(List));
}

Node * node_create(DATA_TYPE k)
{
     Node *node = malloc(sizeof(Node));
     node->data = k;
     node->prev = NULL;
     node->next = NULL;
     return node;
}

void list_destory(List *list)
{
     free(list);
}

void list_print(List* list)
{
     Node *node = list->head;
     while (node != NULL) {
          printf("%d -> ", node->data);
          node = node->next;
     }
     printf("\n");         
}

Node * list_search(List *list, DATA_TYPE k)
{
     Node *node = list->head;
     while (node != NULL && node->data != k)
          node = node->next;
     return node;
}

void list_insert(List *list, Node *x)
{
     x->next = list->head;
     if (list->head != NULL)
          list->head->prev = x;
     list->head = x;
     x->prev = NULL;
}

void list_delete(List *list, Node *x)
{
     if (x->prev != NULL)
          x->prev->next = x->next;
     else
          list->head = x->next;
     if (x->next != NULL)
          x->next->prev = x->prev;
}


源文件 list_test.c:测试数据。

#include <stdio.h>
#include "list.h"

int main(void)
{
     List *list = list_create();

     list_insert(list, node_create(13));
     list_insert(list, node_create(8));
     list_insert(list, node_create(11));
     list_print(list);

     Node *node = list_search(list, 8);
     list_delete(list, node);    
     list_print(list);    

     list_destory(list);

     return 1;
}



posted on 2012-03-05 22:21  毛小娃  阅读(125)  评论(0编辑  收藏  举报

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