基础:通过树的结构实现一个进行字符串反复查询的结构:
208. 实现 Trie (前缀树) https://leetcode-cn.com/problems/implement-trie-prefix-tree/
typedef struct Trie{
struct Trie * children[26];
int isEnd;
} Trie;
/** Initialize your data structure here. */
Trie* trieCreate() {
Trie *node = (Trie *)malloc(sizeof(Trie));
memset(node->children, 0, sizeof(node->children));
node->isEnd = false;
return node;
}
/** Inserts a word into the trie. */
void trieInsert(Trie* obj, char * word) {
int len = strlen(word);
for (int i = 0;i < len;++i) {
int ch = word[i] - 'a';
if (obj->children[ch] == NULL) {
obj->children[ch] = trieCreate();
}
obj = obj->children[ch];
}
obj->isEnd = true;
}
/** Returns if the word is in the trie. */
bool trieSearch(Trie* obj, char * word) {
int len = strlen(word);
for (int i = 0;i < len;++i) {
int ch = word[i] - 'a';
if (obj->children[ch] == NULL) {
return false;
}
obj = obj->children[ch];
}
return obj->isEnd;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
bool trieStartsWith(Trie* obj, char * prefix) {
int len = strlen(prefix);
for (int i = 0;i < len;++i) {
int ch = prefix[i] - 'a';
if (obj->children[ch] == NULL) {
return false;
}
obj = obj->children[ch];
}
return true;
}
void trieFree(Trie* obj) {
for (int i = 0;i < 26;++i) {
if (obj->children[i] != NULL) {
trieFree(obj->children[i]);
}
}
free(obj);
}
/**
* Your Trie struct will be instantiated and called as such:
* Trie* obj = trieCreate();
* trieInsert(obj, word);
* bool param_2 = trieSearch(obj, word);
* bool param_3 = trieStartsWith(obj, prefix);
* trieFree(obj);
*/相关练习:
648. 单词替换 https://leetcode-cn.com/problems/replace-words/
820. 单词的压缩编码 https://leetcode-cn.com/problems/short-encoding-of-words/
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