字典树（Trie树相关）
208. Implement Trie (Prefix Tree)
Implement a trie with insert, search, and startsWith methods. （Medium）
Note:
You may assume that all inputs are consist of lowercase letters a-z.
分析：
字典树即前缀匹配树，在空间不是很影响的情况下一般采用如下数据结构存储Trie节点
class TrieNode {
public:
// Initialize your data structure here.
bool isWord;
TrieNode* child[];
TrieNode(): isWord(false){
memset(child, NULL, sizeof(child));
}
};
所以插入、查找等操作比较直观，直接见程序。
代码：
class TrieNode {
public:
// Initialize your data structure here.
bool isWord;
TrieNode* child[];
TrieNode(): isWord(false){
memset(child, NULL, sizeof(child));
}
};
class Trie {
public:
Trie() {
root = new TrieNode();
}
// Inserts a word into the trie.
void insert(string word) {
TrieNode* r = root;
for (int i = ; i < word.size(); ++i) {
if (r -> child[word[i] - 'a'] == NULL) {
r -> child[word[i] - 'a'] = new TrieNode();
}
r = r -> child[word[i] - 'a'];
}
r -> isWord = true;
}
// Returns if the word is in the trie.
bool search(string word) {
TrieNode* r = root;
for (int i = ; i < word.size(); ++i) {
if (r -> child[word[i] - 'a'] == NULL) {
return false;
}
r = r -> child[word[i] - 'a'];
}
if (r -> isWord) {
return true;
}
return false;
}
// Returns if there is any word in the trie
// that starts with the given prefix.
bool startsWith(string prefix) {
TrieNode* r = root;
for (int i = ; i < prefix.size(); ++i) {
if (r -> child[prefix[i] - 'a'] == NULL) {
return false;
}
r = r -> child[prefix[i] - 'a'];
}
return true;
}
private:
TrieNode* root;
};
// Your Trie object will be instantiated and called as such:
// Trie trie;
// trie.insert("somestring");
// trie.search("key");
211. 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. （Medium）
For 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.
分析：
继续沿用字典树的思路，只不过加入了“.”通配符，所以在查找过程中加入DFS即可，用辅助函数helper
代码：
class TrieNode {
public:
// Initialize your data structure here.
bool isWord;
TrieNode* child[];
TrieNode(): isWord(false){
memset(child, NULL, sizeof(child));
}
};
class Trie {
public:
Trie() {
root = new TrieNode();
}
// Inserts a word into the trie.
void insert(string word) {
TrieNode* r = root;
for (int i = ; i < word.size(); ++i) {
if (r -> child[word[i] - 'a'] == NULL) {
r -> child[word[i] - 'a'] = new TrieNode();
}
r = r -> child[word[i] - 'a'];
}
r -> isWord = true;
}
//helper function for search, use to solve "." problem
bool helper(string word, int pos, TrieNode* curRoot) {
if (pos == word.size() && curRoot -> isWord) {
return true;
}
if (word[pos] != '.') {
if (curRoot -> child[word[pos] - 'a'] == NULL) {
return false;
}
else {
return helper(word, pos + , curRoot -> child[word[pos] - 'a']);
}
}
else {
bool flag = false;
for (int i = ; i < ; ++i) {
if (curRoot -> child[i] != NULL && helper(word, pos + , curRoot -> child[i]) ) {
flag = true;
}
}
return flag;
}
return true;
}
// Returns if the word is in the trie.
bool search(string word) {
return helper(word, , root);
}
private:
TrieNode* root;
};
class WordDictionary {
public:
Trie dic;
// Adds a word into the data structure.
void addWord(string word) {
dic.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) {
return dic.search(word);
}
};
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");