题目链接

写一个数据结构, 支持两种操作。 加入一个字符串, 查找一个字符串是否存在。查找的时候, '.'可以代表任意一个字符。

显然是Trie树, 添加就是正常的添加, 查找的时候只要dfs查找就可以。 具体dfs方法看代码。

struct node
{
node *next[];
int idEnd;
node() {
memset(next, NULL, sizeof(next));
idEnd = ;
}
};
class WordDictionary {
public:
// Adds a word into the data structure.
node *root = new node();
void addWord(string word) {
node *p = root;
int len = word.size();
for(int i = ; i<len; i++) {
if(p->next[word[i]-'a'] == NULL)
p->next[word[i]-'a'] = new node();
p = p->next[word[i]-'a'];
}
p->idEnd = ;
return ;
}
int dfs(string word, int pos, node *p) { //pos是代表当前是在哪一位。
if(pos == word.size())
return p->idEnd;
int len = word.size();
for(int i = pos; i<len; i++) {
if(word[i] == '.') {
for(int j = ; j<; j++) {
if(p->next[j] == NULL)
continue;
if(dfs(word, pos+, p->next[j]))
return ;
}
return ;
} else if(p->next[word[i]-'a'] != NULL) {
return dfs(word, pos+, p->next[word[i]-'a']);
} else {
return ;
}
}
}
// 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 dfs(word, , root);
}
};

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