【刷题-LeetCode】211. Add and Search Word - Data structure design
- Add and Search Word - Data structure design
Design a data structure that supports the following two operations:
void addWord(word)
bool search(word)
search(word) can search a literal word or a regular expression string containing only letters a-z
or .
. A .
means it can represent any one letter.
Example:
addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true
Note:
You may assume that all words are consist of lowercase letters a-z
.
解法1 hashmap。将单词按照长度映射
class WordDictionary {
public:
unordered_map<int, set<string>>mp;
/** Initialize your data structure here. */
WordDictionary() { }
/** Adds a word into the data structure. */
void addWord(string word) {
int m = word.size();
mp[m].insert(word);
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
bool search(string word) {
int m = word.size();
if(mp.count(m) == 0)return false;
for(auto s : mp[m]){
int i;
for(i = 0; i < m; ++i){
if(s[i] == word[i] || word[i] == '.')continue;
else break;
}
if(i == m)return true;
}
return false;
}
};
解法2 trie-tree。查找时,对于包含了.
的部分,递归往后查询
class trie_node{
public:
bool isWord;
vector<trie_node*>child;
trie_node() : isWord(false), child(26, NULL){}
~trie_node(){
for(auto &c : child)delete c;
}
};
class WordDictionary {
public:
trie_node* root;
/** Initialize your data structure here. */
WordDictionary() {
root = new trie_node;
}
/** Adds a word into the data structure. */
void addWord(string s) {
trie_node *cur = root;
for(int i = 0; i < s.size(); ++i){
int idx = s[i] - 'a';
if(cur->child[idx] == NULL){
cur->child[idx] = new trie_node();
}
cur = cur->child[idx];
}
cur->isWord = true;
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
bool search(string word) {
return _search(word, root);
}
bool _search(string s, trie_node* root){
if(s == "")return root->isWord;
if(s[0] == '.'){
for(int j = 0; j < 26; ++j){
if(root->child[j] && _search(s.substr(1), root->child[j]))return true;
}
return false;
}else{
if(root->child[s[0]-'a'] == NULL)return false;
return _search(s.substr(1), root->child[s[0]-'a']);
}
}
};
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