使用 C 链表的 DSA

概述

链接列表是包含项目的链接序列。每个链接都包含到另一个链接的连接。链表是继数组之后第二常用的数据结构。以下是理解链表概念的重要术语。

  • 链接- 链表的每个链接都可以存储称为元素的数据。

  • Next - 链接列表的每个链接都包含一个指向下一个链接的链接，称为“Next”。

  • LinkedList - LinkedList 包含指向名为 First 的第一个链接的连接链接。

链表表示

链表

根据上图所示，以下是需要考虑的要点。

  • LinkedList 包含一个名为first 的链接元素。

  • 每个链接携带一个数据字段和一个接下来调用的链接字段。

  • 每个链接都使用其下一个链接与其下一个链接进行链接。

  • Last Link 带有一个为 null 的 Link 来标记列表的结尾。

链表的类型

以下是链表的各种风格。

  • 简单链接列表- 项目导航仅向前。

  • 双向链表- 项目可以向前和向后导航。

  • 循环链表- 最后一项包含第一个元素作为下一个元素的链接，第一个元素包含到最后一个元素作为上一个元素的链接。

基本操作

以下是列表支持的基本操作。

  • 插入- 在列表的开头添加一个元素。

  • 删除- 删除列表开头的元素。

  • 显示- 显示完整列表。

  • 搜索- 使用给定键搜索元素。

  • 删除- 使用给定键删除元素。

插入操作

插入是一个三步过程 -

  • 使用提供的数据创建新链接。

  • 将新链接指向旧的第一个链接。

  • 将第一个链接指向此新链接。

链表先插入 

//insert link at the first location
void insertFirst(int key, int data){
   //create a link
   struct node *link = (struct node*) malloc(sizeof(struct node));
   link->key = key;
   link->data = data;
   
   //point it to old first node
   link->next = head;
   
   //point first to new first node
   head = link;
}

删除操作

删除是一个两步过程 -

  • 获取第一个链接指向的链接作为临时链接。

  • 将第一个链接指向临时链接的下一个链接。

链表先删除 

//delete first item
struct node* deleteFirst(){
   //save reference to first link
   struct node *tempLink = head;
   
   //mark next to first link as first 
   head = head->next;
   
   //return the deleted link
   return tempLink;
}

导航操作

导航是一个递归步骤过程，是搜索、删除等许多操作的基础。 -

  • 获取第一个链接指向的链接作为当前链接。

  • 检查当前链接是否不为空并显示它。

  • 将当前链接指向当前链接的下一个链接并移至上述步骤。

链表导航

注意 -

//display the list
void printList(){
   struct node *ptr = head;
   printf("\n[ ");
   
   //start from the beginning
   while(ptr != NULL){        
      printf("(%d,%d) ",ptr->key,ptr->data);
      ptr = ptr->next;
   }
   printf(" ]");
}

高级操作

以下是为列表指定的高级操作。

  • 排序- 根据特定顺序对列表进行排序。

  • 反转- 反转链表。

排序操作

我们使用冒泡排序来对列表进行排序。

void sort(){
   int i, j, k, tempKey, tempData ;
   struct node *current;
   struct node *next;
   int size = length();
   k = size ;
   for ( i = 0 ; i < size - 1 ; i++, k-- ) {
      current = head ;
      next = head->next ;
      for ( j = 1 ; j < k ; j++ ) {            
         if ( current->data > next->data ) {
            tempData = current->data ;
            current->data = next->data;
            next->data = tempData ;

            tempKey = current->key;
            current->key = next->key;
            next->key = tempKey;
         }
         current = current->next;
         next = next->next;                        
      }
   }
}

反向操作

以下代码演示了反转单个链表。

void reverse(struct node** head_ref) {
   struct node* prev   = NULL;
   struct node* current = *head_ref;
   struct node* next;
   while (current != NULL) {
      next  = current->next;  
      current->next = prev;   
      prev = current;
      current = next;
   }
   *head_ref = prev;
}

例子

LinkedListDemo.c

现场演示 
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>

struct node {
   int data;
   int key;
   struct node *next;
};
struct node *head = NULL;
struct node *current = NULL;

//display the list
void printList(){
   struct node *ptr = head;
   printf("\n[ ");
   
   //start from the beginning
   while(ptr != NULL){        
      printf("(%d,%d) ",ptr->key,ptr->data);
      ptr = ptr->next;
   }
   printf(" ]");
}
//insert link at the first location
void insertFirst(int key, int data){
   //create a link
   struct node *link = (struct node*) malloc(sizeof(struct node));
   link->key = key;
   link->data = data;
   
   //point it to old first node
   link->next = head;
   
   //point first to new first node
   head = link;
}
//delete first item
struct node* deleteFirst(){
   //save reference to first link
   struct node *tempLink = head;
   
   //mark next to first link as first 
   head = head->next;
   
   //return the deleted link
   return tempLink;
}
//is list empty
bool isEmpty(){
   return head == NULL;
}
int length(){
   int length = 0;
   struct node *current;
   for(current = head; current!=NULL;
      current = current->next){
      length++;
   }
   return length;
}
//find a link with given key
struct node* find(int key){
   //start from the first link
   struct node* current = head;

   //if list is empty
   if(head == NULL){
      return NULL;
   }
   //navigate through list
   while(current->key != key){
      //if it is last node
      if(current->next == NULL){
         return NULL;
      } else {
         //go to next link
         current = current->next;
      }
   }
   //if data found, return the current Link
   return current;
}
//delete a link with given key
struct node* delete(int key){
   //start from the first link
   struct node* current = head;
   struct node* previous = NULL;
   
   //if list is empty
   if(head == NULL){
      return NULL;
   }
   //navigate through list
   while(current->key != key){
      //if it is last node
      if(current->next == NULL){
         return NULL;
      } else {
         //store reference to current link
         previous = current;
         
         //move to next link
         current = current->next;             
      }
   }
   //found a match, update the link
   if(current == head) {
      //change first to point to next link
      head = head->next;
   } else {
      //bypass the current link
      previous->next = current->next;
   }
   return current;
}
void sort(){
   int i, j, k, tempKey, tempData ;
   struct node *current;
   struct node *next;
   int size = length();
   k = size ;
   for ( i = 0 ; i < size - 1 ; i++, k-- ) {
      current = head ;
      next = head->next ;
      for ( j = 1 ; j < k ; j++ ) {            
         if ( current->data > next->data ) {
            tempData = current->data ;
            current->data = next->data;
            next->data = tempData ;

            tempKey = current->key;
            current->key = next->key;
            next->key = tempKey;
         }
         current = current->next;
         next = next->next;                        
      }
   }
}
void reverse(struct node** head_ref) {
   struct node* prev   = NULL;
   struct node* current = *head_ref;
   struct node* next;
   while (current != NULL) {
      next  = current->next;  
      current->next = prev;   
      prev = current;
      current = next;
   }
   *head_ref = prev;
}

main() {
   insertFirst(1,10);
   insertFirst(2,20);
   insertFirst(3,30);
   insertFirst(4,1);
   insertFirst(5,40);
   insertFirst(6,56); 

   printf("Original List: "); 
   //print list
   printList();

   while(!isEmpty()){            
      struct node *temp = deleteFirst();
      printf("\nDeleted value:");  
      printf("(%d,%d) ",temp->key,temp->data);        
   }         
   printf("\nList after deleting all items: ");          
   printList();
   insertFirst(1,10);
   insertFirst(2,20);
   insertFirst(3,30);
   insertFirst(4,1);
   insertFirst(5,40);
   insertFirst(6,56); 
   printf("\nRestored List: ");  
   printList();
   printf("\n");  

   struct node *foundLink = find(4);
   if(foundLink != NULL){
      printf("Element found: ");  
      printf("(%d,%d) ",foundLink->key,foundLink->data);  
      printf("\n");  
   } else {
      printf("Element not found.");  
   }
   delete(4);
   printf("List after deleting an item: ");  
   printList();
   printf("\n");
   foundLink = find(4);
   if(foundLink != NULL){
      printf("Element found: ");  
      printf("(%d,%d) ",foundLink->key,foundLink->data);  
      printf("\n");  
   } else {
      printf("Element not found.");  
   }
   printf("\n");  
   sort();
   printf("List after sorting the data: ");  
   printList();
   reverse(&head);
   printf("\nList after reversing the data: ");  
   printList();
}

输出

如果我们编译并运行上面的程序，那么它将产生以下输出 -

Original List: 
[ (6,56) (5,40) (4,1) (3,30) (2,20) (1,10) ]
Deleted value:(6,56) 
Deleted value:(5,40) 
Deleted value:(4,1) 
Deleted value:(3,30) 
Deleted value:(2,20) 
Deleted value:(1,10) 
List after deleting all items: 
[ ]
Restored List: 
[ (6,56) (5,40) (4,1) (3,30) (2,20) (1,10) ]
Element found: (4,1) 
List after deleting an item: 
[ (6,56) (5,40) (3,30) (2,20) (1,10) ]
Element not found.
List after sorting the data: 
[ (1,10) (2,20) (3,30) (5,40) (6,56) ]
List after reversing the data: 
[ (6,56) (5,40) (3,30) (2,20) (1,10) ]