week6 solution add

Author: hoyeongkwak

design-add-and-search-words-data-structure/hoyeongkwak.java

@@ -0,0 +1,80 @@
+/*
+m : word length
+n : Trie node count
+addWord
+Time Complexity: O(m)
+
+search
+Time Complexity: O(n)
+
+Space Complexity: O(26 × N × M)
+
+Trie + Dfs
+*/
+
+class WordDictionary {
+    class TrieNode {
+        TrieNode[] children;
+        boolean isEnd;
+
+        public TrieNode() {
+            children = new TrieNode[26];
+            isEnd = false;
+        }
+    }
+
+    private TrieNode root;
+
+    public WordDictionary() {
+        root = new TrieNode();
+    }
+    
+    public void addWord(String word) {
+        TrieNode current = root;
+
+        for (char str : word.toCharArray()) {
+            int idx = str - 'a';
+            if (current.children[idx] == null) {
+                current.children[idx] = new TrieNode();
+            }
+            current = current.children[idx];
+        }
+        current.isEnd = true;
+    }
+    
+    public boolean search(String word) {
+        return dfsSearch(word, 0, root);
+    }
+
+    private boolean dfsSearch(String word, int idx, TrieNode node) {
+        if (idx == word.length()) {
+            return node.isEnd;
+        }
+
+        char c = word.charAt(idx);
+        if (c == '.') {
+            for (int i = 0 ; i < 26; i++) {
+                if (node.children[i] != null) {
+                    if (dfsSearch(word, idx + 1, node.children[i])) {
+                        return true;
+                    }
+                }
+            }
+            return false;
+        } else {
+            int charIdx = c - 'a';
+            if (node.children[charIdx] == null) {
+                return false;
+            }
+            return dfsSearch(word, idx + 1, node.children[charIdx]);
+        }
+    }
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * WordDictionary obj = new WordDictionary();
+ * obj.addWord(word);
+ * boolean param_2 = obj.search(word);
+ */
+