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循环单链表在C语言中的优雅实现?

singly-linked-list linked-list data-structures c

matcheek · 技术社区 · 14 年前

经过经典的数据结构,在链表上停下来了,刚实现了一个循环的单链表,但我印象深刻的是,这个链表可以用更优雅的方式表达,特别是删除节点功能。考虑到效率和代码可读性,有人能为单独链接的循环列表提供一个更简洁和高效的解决方案吗?

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


struct node{
    struct node* next;
    int value;
};


struct list{
    struct node* head;
};


struct node* init_node(int value){
    struct node* pnode;
    if (!(pnode = (struct node*)malloc(sizeof(struct node)))){
        return NULL;
    }
    else{
        pnode->value = value;   
    }
    return pnode;
}

struct list* init_list(){
    struct list* plist;
    if (!(plist = (struct list*)malloc(sizeof(struct list)))){
        return NULL;        
    }
    plist->head = NULL;
    return plist;
}


void remove_node(struct list*a plist, int value){

    struct node* current, *temp;
    current = plist->head;
    if (!(current)) return; 
    if ( current->value == value ){
        if (current==current->next){
            plist->head = NULL; 
            free(current);
        }
        else {
            temp = current;
            do {    
                current = current->next;    
            } while (current->next != plist->head);

            current->next = plist->head->next;
            plist->head = current->next;
            free(temp);
        }
    }
    else {
        do {
            if (current->next->value == value){
                temp = current->next;
                current->next = current->next->next;
                free(temp);
            }
            current = current->next;
        } while (current != plist->head);
    }
}

void print_node(struct node* pnode){
    printf("%d %p %p\n", pnode->value, pnode, pnode->next); 
}
void print_list(struct list* plist){

    struct node * current = plist->head;

    if (!(current)) return;
    if (current == plist->head->next){
        print_node(current);
    }
    else{
        do {
            print_node(current);
            current = current->next;

        } while (current != plist->head);
    }

}

void add_node(struct node* pnode,struct list* plist){

    struct node* current;
    struct node* temp;
    if (plist->head == NULL){
        plist->head = pnode;
        plist->head->next = pnode;
    }
    else {
        current = plist->head;
        if (current == plist->head->next){
            plist->head->next = pnode;
            pnode->next = plist->head;      
        }
        else {
            while(current->next!=plist->head)
                current = current->next;

            current->next = pnode;
            pnode->next = plist->head;
        }

    }
}

4 回复 | 直到 8 年前

pitti 14 年前

http://lxr.linux.no/linux+v2.6.36/include/linux/list.h

struct list_head * struct list_head

Fabian Giesen 14 年前

struct node* get_sentinel(struct list* plist)
{
    // use &plist->head itself as sentinel!
    // (works because struct node starts with the next pointer)
    return (struct node*) &plist->head;
}

struct list* init_list(){
    struct list* plist;
    if (!(plist = (struct list*)malloc(sizeof(struct list)))){
        return NULL;        
    }
    plist->head = get_sentinel(plist);
    return plist;
}

void add_node_at_front(struct node* pnode,struct list* plist){
    pnode->next = plist->head;
    plist->head = pnode;
}

void add_node_at_back(struct node* pnode,struct list* plist){
    struct node *current, *sentinel = get_sentinel(plist);

    // search for last element
    current = plist->head;
    while (current->next != sentinel)
        current = current->next;

    // insert node
    pnode->next = sentinel;
    current->next = pnode;
}

void remove_node(struct list* plist, int value){
    struct node **prevnext, *sentinel = get_sentinel(plist);
    prevnext = &plist->head; // ptr to next pointer of previous node
    while (*prevnext != sentinel) {
        struct node *current = *prevnext;
        if (current->value == value) {
            *prevnext = current->next; // remove current from list
            free(current); // and free it
            break; // we're done!
        }
        prevnext = &current->next;
    }
}

void print_list(struct list* plist){
    struct node *current, *sentinel = get_sentinel(plist);
    for (current = plist->head; current != sentinel; current = current->next)
        print_node(current);
}

uesp 14 年前

KeyC0de 8 年前

Node* createCircularLList(int size)
{
    Node *it; // to iterate through the LList
    Node *head;

    // Create the head /1st Node of the list
    head = it = (Node*)malloc(sizeof(Node));
    head->id = 1;

    // Create the remaining Nodes (from 2 to size)
    int i;
    for (i = 2; i <= size; ++i) {
        it->next = (Node*)malloc(sizeof(Node)); // create next Node
        it = it->next;                          // point to it
        it->id = i;                             // assign its value / id
        if (i == 2)
            head->next = it; // head Node points to the 2nd Node
    }
    // close the llist by having the last Node point to the head Node
    it->next = head;

    return head;    // return pointer to start of the list
}

Node

typedef struct Node {
    int id;
    struct Node *next;
} Node;