design add and search words data structure solution

design-add-and-search-words-data-structure/jaejeong1.java

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+import java.util.HashMap;
+import java.util.Map;
+
+class WordDictionary {
+  // 시간복잡도: O(N), N(word의 길이)
+  // 공간복잡도: O(N)
+  Map<Character, WordDictionary> child;
+  boolean isEnd;
+
+  public WordDictionary() {
+    child = new HashMap<>();
+    isEnd = false;
+  }
+
+  public void addWord(String word) {
+    var node = this;
+    for (int i = 0; i < word.length(); i++) {
+      char c = word.charAt(i);
+
+      node.child.putIfAbsent(c, new WordDictionary());
+      node = node.child.get(c);
+
+      if (i == word.length() - 1) {
+        node.isEnd = true;
+        return;
+      }
+    }
+  }
+
+  public boolean search(String word) {
+    return searchHelper(word, 0, this);
+  }
+
+  private boolean searchHelper(String word, int index, WordDictionary node) {
+    if (index == word.length()) {
+      return node.isEnd;
+    }
+
+    char c = word.charAt(index);
+    // . 이 나오면 해당 노드의 자식 노드들에 대해 모두 재귀를 돌린다
+    if (c == '.') {
+      for (WordDictionary childNode : node.child.values()) {
+        if (searchHelper(word, index + 1, childNode)) {
+          return true;
+        }
+      }
+      return false;
+    } else {
+      var childNode = node.child.get(c);
+      if (childNode == null) {
+        return false;
+      }
+      return searchHelper(word, index + 1, childNode);
+    }
+  }
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * WordDictionary obj = new WordDictionary();
+ * obj.addWord(word);
+ * boolean param_2 = obj.search(word);
+ */