From 606209d7cfccdc77aeac61d507e03c7241a70198 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Tue, 7 Jul 2020 20:30:12 +0800
Subject: [PATCH] Create 211. Add and Search Word - Data structure design.md

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 ...and Search Word - Data structure design.md | 59 +++++++++++++++++++
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 create mode 100644 solutions/211. Add and Search Word - Data structure design.md

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+# [211. Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design/)
+
+# 思路
+
+ 这题其实就是[208. Implement Trie(Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree/)的升级版，会做208题此题就没啥问题了。
+ 不同的地方就是查询时`.`可以代替任意字符，所以一旦有了`.`，就需要查找所有的子树，典型的DFS的问题。
+ 
+ # C++
+ ```C++
+class TrieNode{
+public:
+    bool isLeaf;
+    vector<TrieNode *>nexts = vector<TrieNode *>(26, NULL);
+    TrieNode(bool _isLeaf){
+        isLeaf = _isLeaf;
+    } 
+};
+
+class WordDictionary {
+private:
+    TrieNode *root = new TrieNode(false);
+public:
+    /** Initialize your data structure here. */
+    WordDictionary() {}
+    
+    /** Adds a word into the data structure. */
+    void addWord(string word) {
+        TrieNode *p = root;
+        for(char c: word){
+            if(p -> nexts[c-'a'] == NULL){
+                TrieNode *node = new TrieNode(false);
+                p -> nexts[c-'a'] = node;
+                p = node;
+            }
+            else p = p -> nexts[c-'a'];
+        }
+        p -> isLeaf = true;
+    }
+    
+    bool DFS(const string &word, TrieNode *node, int start){
+        if(start == word.size()) return node -> isLeaf;
+        
+        if(word[start] == '.'){
+            for(TrieNode *p: node -> nexts)
+                if(p != NULL && DFS(word, p, start+1)) return true;
+            return false;
+        }
+        
+        TrieNode *p = node -> nexts[word[start] - 'a'];
+        if(!p) return false;
+        return DFS(word, p, start + 1);
+    }
+    
+    /** 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, 0);
+    }
+};
+ ```