[英]Elegant implementation of circular singly-linked list in C?
經歷了經典的數據結構並停止在鏈表上。只是實現了一個循環的單鏈表,但我印象深刻的是這個列表可以用更優雅的方式表達,特別是remove_node函數。 牢記效率和代碼可讀性,任何人都可以為單鏈循環列表提供更簡潔有效的解決方案嗎?
#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;
}
}
}
看一下Linux內核源代碼中的循環鏈表: http : //lxr.linux.no/linux+v2.6.36/include/linux/list.h
它的美妙之處在於你沒有一個特殊的結構來使你的數據適合列表,你只需要在你希望作為列表的結構中包含struct list_head *
。 用於訪問列表中項目的宏將處理偏移量計算,以從struct list_head
指針獲取數據。
可以在kernelnewbies.org/FAQ/LinkedLists上找到對內核中使用的鏈表的更詳細說明(對不起,我沒有足夠的業力來發布兩個超鏈接)。
編輯:嗯,列表是一個雙鏈表,而不是像你一樣的單鏈表,但你可以采用這個概念並創建自己的單鏈表。
當您將列表頭部視為列表的元素(所謂的“哨兵”)時,列表處理(特別是循環列表)變得更容易。 很多特殊情況都消失了。 您可以為sentinel使用虛擬節點,但如果下一個指針在結構中是第一個,則您甚至不需要這樣做。 另一個重要的技巧是在修改列表時保持指向前一個節點的下一個指針(以便稍后可以修改它)。 總而言之,你得到這個:
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 = ¤t->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);
}
一些評論:
我使用以下內容創建一個動態循環單鏈表。 所需要的只是尺寸。
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
}
我定義Node
ADT如下:
typedef struct Node {
int id;
struct Node *next;
} Node;
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