散列技术之链地址法(基于无序链表)

源码例如以下：

#include <stdlib.h>
#include <stdio.h>
#define hash(v,M) (v % M)
typedef char Key;
struct Item{
	Key key;
};
typedef struct STnode* link;
struct STnode{
	Item item ; link next;
};


static link* heads , z ;
static struct Item NULLitem ;
static int N , M ;


Key key(Item item){
	return item.key;
}
//创建一个节点 
static link NEW(Item item, link next){
	link x = (link)malloc(sizeof *x);
	x->item = item;x->next = next;
	return x;
}
//初始化 
void STinit(int max){
	int i ;
	N = 0; M = max / 5;
	heads = (link *)malloc(M*sizeof(link));
	z = NEW(NULLitem,NULL);
	for(i=0;i<M;i++)heads[i] = z;
}
//节点个数 
int STcount(){
	return N; 
} 
//搜索子程序 
Item searchR(link h, Key v){
	Key t = key(h->item);
	if(h==z)return NULLitem;
	if(v==t) return h->item;
	return searchR(h->next,v);
}
//搜索主程序 
Item STsearch(Key v){
	return searchR(heads[hash(v,M)],v);
}


//插入主程序 
void STinsert(Item item){
	int i = hash(key(item),M) ;
	heads[i] = NEW(item,heads[i]); N++;
}


//单链表删除指定的节点  非递归 
link deleteT(link head,Item v){
	link node1=head;
	link node2=NULL;
	if (head==NULL)return NULL;
	if (node1->item.key==v.key){
		head=head->next;
		free(node1);
		return head;
	}
	while (node1!=NULL){
		node2=node1;
		node2=node2->next;
		if (node2->item.key==v.key){
			node1->next=node2->next;
			free(node2);
			break;
		}
		node1=node1->next;
	}
	return head;
}




//单链表删除指定的节点  递归 
link deleteR(link h,Item item){
	if(h==NULL)return NULL;
	if(key(h->item) == key(item))return h->next;
	h->next = deleteR(h->next,item);
	return h;
}


void STdelete(Item item){
	int i = hash(key(item),M) ;
	heads[i] = deleteR(heads[i],item);
}


void p(link h){
	link t = h;
	while(t!=NULL){
		printf("%c ",key(t->item));
		t = t->next;
	}
	printf("\n");
	
}
main(){
	STinit(26);
	struct Item item[13] ={'a','s','e','r','c','h','i','n','g','x','m','p','l'};
	int i;
	for(i = 0; i<13;i++)
		STinsert(item[i]);
	for(i = 0; i<M;i++)
		p(heads[i]);
	printf("search: %c \n",key(STsearch('e')));
	struct Item delItem = {'a'};
	STdelete(delItem);
	for(i = 0; i<M;i++)
		p(heads[i]);
	
	
}