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.

For example:

search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

You may assume that all words are consist of lowercase letters a-z.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.


LeetCode出新题的速度越来越快了,有点跟不上节奏的感觉了。这道题如果做过之前的那道 Implement Trie (Prefix Tree) 实现字典树(前缀树)的话就没有太大的难度了,还是要用到字典树的结构,唯一不同的地方就是search的函数需要重新写一下,因为这道题里面’.’可以代替任意字符,所以一旦有了’.’,就需要查找所有的子树,只要有一个返回true,整个search函数就返回true,典型的DFS的问题,其他部分跟上一道实现字典树没有太大区别,代码如下:


class WordDictionary {
    struct TrieNode {
        TrieNode *child[26];
        bool isWord;
        TrieNode() : isWord(false) {
            for (auto &a : child) a = NULL;
    WordDictionary() {
        root = new TrieNode();
    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode *p = root;
        for (auto &a : word) {
            int i = a - 'a';
            if (!p->child[i]) p->child[i] = new TrieNode();
            p = p->child[i];
        p->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 searchWord(word, root, 0);
    bool searchWord(string &word, TrieNode *p, int i) {
        if (i == word.size()) return p->isWord;
        if (word[i] == '.') {
            for (auto &a : p->child) {
                if (a && searchWord(word, a, i + 1)) return true;
            return false;
        } else {
            return p->child[word[i] - 'a'] && searchWord(word, p->child[word[i] - 'a'], i + 1);
    TrieNode *root;

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");


讨论:这道题有个很好的Follow up,就是当搜索的单词中存在星号怎么搞,星号的定义和Wildcard Matching中一样,可以代表任意的字符串,包括空字符串,请参见评论区1楼。



Implement Trie (Prefix Tree)

Wildcard Matching





LeetCode All in One 题目讲解汇总(持续更新中…)

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