LeetCode-211 Add and Search Word

题目描述

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.

题目大意

实现两个操作：插入单词、查找单词（查找单词时的输入只能为 'a-z' 和 '.' ，其中 '.' 可以代表任何小写字母）。

示例

E

解题思路

解题思路类似于LeetCode - 208 Implement Trie，设置一个node class用来记录保存树结点中的节点信息。

插入单词是建树的过程，查找单词是搜索树的过程。

复杂度分析

时间复杂度：O(N)

空间复杂度：O(N)

代码

class TrieNode {
public:
    //该节点代表的字母是否是某个单词的最后一个字母
    bool word;
    TrieNode* children[26];
    TrieNode() {
        word = false;
        memset(children, NULL, sizeof(children));
    }
};

class WordDictionary {
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        
    }
    
    //建立一棵树
    /** Adds a word into the data structure. */
    void addWord(string word) {
        TrieNode* node = root;
        for (char c : word) {
            if (!node -> children[c - 'a']) {
                node -> children[c - 'a'] = new TrieNode();
            }
            node = node -> children[c - 'a'];
        }
        node -> word = 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.c_str(), root);
    }
private:
    TrieNode* root = new TrieNode();
    
    bool search(const char* word, TrieNode* node) {
        for (int i = 0; word[i] && node; i++) {
            //若搜索的单词该位置不是‘.’，则直接进行比较，否则对所有子结点进行遍历搜索
            if (word[i] != '.') {
                node = node -> children[word[i] - 'a'];
            } else {
                TrieNode* tmp = node;
                for (int j = 0; j < 26; j++) {
                    node = tmp -> children[j];
                    if (search(word + i + 1, node)) {
                        return true;
                    }
                }
            }
        }
        return node && node -> word;
    }
};