您可以使用單個二叉查找樹(AVL/Red-black/...)來包含所有集合中的字符串全部,方法是按字典順序將它們鍵入爲(set_number,string)。你不需要在任何地方明確地存儲集合。例如,定義節點的順序爲樹比較可能看起來像:
function compare_nodes (node1, node2) {
if (node1.set_number < node2.set_number) return LESS;
if (node1.set_number > node2.set_number) return GREATER;
if (node1.string < node2.string) return LESS;
if (node1.string > node2.string) return GREATER;
return EQUAL;
}
通過這樣的結構,一些常用的操作是可能的(但也許並不簡單)。
要查找集合set_number
中是否存在字符串s
,只需在樹中查找(set_number
,s
)即可完全匹配。
要查找一組set_number
所有字符串:以上
function iterate_all_strings_in_set (set_number) {
// Traverse the tree from root downwards, looking for the given key. Return
// wherever the search ends up, whether it found the value or not.
node = lookup_tree_weak(set_number, "");
// tree empty?
if (node == null) {
return;
}
// We may have gotten the greatest node from the previous set,
// instead of the first node from the set we're interested in.
if (node.set_number != set_number) {
node = successor(node);
}
while (node != null && node.set_number == set_number) {
do_something_with(node.string);
node = successor(node);
}
}
需要O((k+1)*log(n))
時間,其中k
是set_number
串的數量,並且n
是所有字符串的數量。
要查找所有組數字與相關聯的至少一個字符串:
function iterate_all_sets()
{
node = first_node_in_tree();
while (node != null) {
current_set = node.set_number;
do_something_with(current_set);
if (cannot increment current_set) {
return;
}
node = lookup_tree_weak(current_set + 1, "");
if (node.set_number == current_set) {
node = successor(node);
}
}
}
上述要求O((k+1)*log(n))
時間,其中k
是套有至少一個串號,並n
是所有字符串的數量。
請注意,上面的代碼假定樹在「do_something」調用中未被修改;如果節點被移除,它可能會崩潰。
Addidionally,這是一些真正的C代碼,它演示了這一點,使用my own generic AVL tree implemetation。爲了編譯它,只需從BadVPN源文件夾中複製misc/
和structure/
文件夾並在其中添加包含路徑就足夠了。
請注意我的AVL樹在其節點中如何不包含任何「數據」,以及它如何不執行任何自己的內存分配。當您需要處理大量數據時,這非常方便。爲了說清楚:下面的程序只有一個malloc()
,它是分配節點數組的節點。
#include <stdlib.h>
#include <stdio.h>
#include <inttypes.h>
#include <assert.h>
#include <structure/BAVL.h>
#include <misc/offset.h>
struct value {
uint32_t set_no;
char str[3];
};
struct node {
uint8_t is_used;
struct value val;
BAVLNode tree_node;
};
BAVL tree;
static int value_comparator (void *unused, void *vv1, void *vv2)
{
struct value *v1 = vv1;
struct value *v2 = vv2;
if (v1->set_no < v2->set_no) {
return -1;
}
if (v1->set_no > v2->set_no) {
return 1;
}
int c = strcmp(v1->str, v2->str);
if (c < 0) {
return -1;
}
if (c > 0) {
return 1;
}
return 0;
}
static void random_bytes (unsigned char *out, size_t n)
{
while (n > 0) {
*out = rand();
out++;
n--;
}
}
static void random_value (struct value *out)
{
random_bytes((unsigned char *)&out->set_no, sizeof(out->set_no));
for (size_t i = 0; i < sizeof(out->str) - 1; i++) {
out->str[i] = (uint8_t)32 + (rand() % 94);
}
out->str[sizeof(out->str) - 1] = '\0';
}
static struct node * find_node (const struct value *val)
{
// find AVL tree node with an equal value
BAVLNode *tn = BAVL_LookupExact(&tree, (void *)val);
if (!tn) {
return NULL;
}
// get node pointer from pointer to its value (same as container_of() in Linux kernel)
struct node *n = UPPER_OBJECT(tn, struct node, tree_node);
assert(n->val.set_no == val->set_no);
assert(!strcmp(n->val.str, val->str));
return n;
}
static struct node * lookup_weak (const struct value *v)
{
BAVLNode *tn = BAVL_Lookup(&tree, (void *)v);
if (!tn) {
return NULL;
}
return UPPER_OBJECT(tn, struct node, tree_node);
}
static struct node * first_node (void)
{
BAVLNode *tn = BAVL_GetFirst(&tree);
if (!tn) {
return NULL;
}
return UPPER_OBJECT(tn, struct node, tree_node);
}
static struct node * next_node (struct node *node)
{
BAVLNode *tn = BAVL_GetNext(&tree, &node->tree_node);
if (!tn) {
return NULL;
}
return UPPER_OBJECT(tn, struct node, tree_node);
}
size_t num_found;
static void iterate_all_strings_in_set (uint32_t set_no)
{
struct value v;
v.set_no = set_no;
v.str[0] = '\0';
struct node *n = lookup_weak(&v);
if (!n) {
return;
}
if (n->val.set_no != set_no) {
n = next_node(n);
}
while (n && n->val.set_no == set_no) {
num_found++; // "do_something_with_string"
n = next_node(n);
}
}
static void iterate_all_sets (void)
{
struct node *node = first_node();
while (node) {
uint32_t current_set = node->val.set_no;
iterate_all_strings_in_set(current_set); // "do_something_with_set"
if (current_set == UINT32_MAX) {
return;
}
struct value v;
v.set_no = current_set + 1;
v.str[0] = '\0';
node = lookup_weak(&v);
if (node->val.set_no == current_set) {
node = next_node(node);
}
}
}
int main (int argc, char *argv[])
{
size_t num_nodes = 10000000;
// init AVL tree, using:
// key=(struct node).val,
// comparator=value_comparator
BAVL_Init(&tree, OFFSET_DIFF(struct node, val, tree_node), value_comparator, NULL);
printf("Allocating...\n");
// allocate nodes (missing overflow check...)
struct node *nodes = malloc(num_nodes * sizeof(nodes[0]));
if (!nodes) {
printf("malloc failed!\n");
return 1;
}
printf("Inserting %zu nodes...\n", num_nodes);
size_t num_inserted = 0;
// insert nodes, giving them random values
for (size_t i = 0; i < num_nodes; i++) {
struct node *n = &nodes[i];
// choose random set number and string
random_value(&n->val);
// try inserting into AVL tree
if (!BAVL_Insert(&tree, &n->tree_node, NULL)) {
printf("Insert collision: (%"PRIu32", '%s') already exists!\n", n->val.set_no, n->val.str);
n->is_used = 0;
continue;
}
n->is_used = 1;
num_inserted++;
}
printf("Looking up...\n");
// lookup all those values
for (size_t i = 0; i < num_nodes; i++) {
struct node *n = &nodes[i];
struct node *lookup_n = find_node(&n->val);
if (n->is_used) { // this node is the only one with this value
ASSERT(lookup_n == n)
} else { // this node was an insert collision; some other
// node must have this value
ASSERT(lookup_n != NULL)
ASSERT(lookup_n != n)
}
}
printf("Iterating by sets...\n");
num_found = 0;
iterate_all_sets();
ASSERT(num_found == num_inserted)
printf("Removing all strings...\n");
for (size_t i = 0; i < num_nodes; i++) {
struct node *n = &nodes[i];
if (!n->is_used) { // must not remove it it wasn't inserted
continue;
}
BAVL_Remove(&tree, &n->tree_node);
}
return 0;
}
多少?它適合內存嗎?爲什麼字符串表示爲鏈接列表?一個字符串是否可以出現在多個列表中?一個列表中有多少個字符串?你已經有了某種形式的數據(內存,磁盤文件)嗎? – wildplasser
你顯然沒有讀過我的問題。我說 - 「我目前的解決方案...」,並開始什麼是我的**想法,在那裏我把這組字符串表示爲一個鏈表。你可能有更好的解決方案?我都聽過這個。是的,一個字符串可能出現在多個列表中。 – user866098