word dictionary solution

Author: jinvicky

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

@@ -0,0 +1,44 @@
+class WordDictionary {
+
+    private static class Node {
+        Node[] next = new Node[26];
+        boolean isEnd;
+    }
+
+    private final Node root = new Node();
+
+    public WordDictionary() {}
+
+    public void addWord(String word) {
+        Node cur = root;
+        for (int k = 0; k < word.length(); k++) {
+            char ch = word.charAt(k);
+            int i = ch - 'a';
+            if (cur.next[i] == null) cur.next[i] = new Node();
+            cur = cur.next[i];
+        }
+        cur.isEnd = true;
+    }
+
+    public boolean search(String word) {
+        return dfs(word, 0, root);
+    }
+
+    private boolean dfs(String word, int idx, Node node) {
+        if (node == null) return false;
+        if (idx == word.length()) return node.isEnd;
+
+        char ch = word.charAt(idx);
+        if (ch == '.') {
+            // 모든 가능 문자로 한 글자 매칭
+            for (int c = 0; c < 26; c++) {
+                if (node.next[c] != null && dfs(word, idx + 1, node.next[c])) {
+                    return true;
+                }
+            }
+            return false;
+        } else {
+            return dfs(word, idx + 1, node.next[ch - 'a']);
+        }
+    }
+}