#290 Word Search II

Author: forest000014

word-search-ii/forest000014.java

@@ -0,0 +1,73 @@
+/*
+# Time Complexity: O(w + m * n * 4^10)
+  - trie에 word 하나(최대 길이 10)를 삽입하는 데에는 O(10) = O(1) 이므로, trie 전체를 생성하는 데에는 O(w) (w는 words의 length)
+  - dfs 탐색을 하면서, 모든 경로를 탐색 (최대 depth는 10)
+# Space Complexity: O(w)
+  - trie를 생성하면, 최대 10글자 * w개 문자열 = O(10w) = O(w)
+*/
+class Solution {
+
+    private class Trie {
+        char val;
+        boolean ends;
+        Map<Character, Trie> children;
+
+        Trie(char val) {
+            this.val = val;
+            this.children = new HashMap<>();
+        }
+    }
+    public List<String> findWords(char[][] board, String[] words) {
+        // Trie 생성 및 세팅
+        Trie root = new Trie('.');
+        for (String word : words) {
+            Trie curr = root;
+            for (int i = 0; i < word.length(); i++) {
+                char ch = word.charAt(i);
+                curr.children.putIfAbsent(ch, new Trie(ch));
+                curr = curr.children.get(ch);
+            }
+            curr.ends = true;
+        }
+
+        // trie와 dfs를 사용하여, 단어가 존재하는지 확인
+        int m = board.length;
+        int n = board[0].length;
+        boolean[][] visited = new boolean[m][n];
+        StringBuilder sb = new StringBuilder();
+        Set<String> ans = new HashSet<>();
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (!root.children.containsKey(board[i][j])) continue;
+
+                visited[i][j] = true;
+                sb.append(board[i][j]);
+                dfs(m, n, board, visited, i, j, root.children.get(board[i][j]), sb, ans); //
+                sb.deleteCharAt(0);
+                visited[i][j] = false;
+            }
+        }
+
+        return ans.stream().collect(Collectors.toList());
+    }
+
+    private void dfs(int m, int n, char[][] board, boolean[][] visited, int r, int c, Trie curr, StringBuilder sb, Set<String> ans) {
+        if (curr.ends) ans.add(sb.toString());
+
+        int[] dr = {-1, 0, 1, 0};
+        int[] dc = {0, 1, 0, -1};
+
+        for (int i = 0; i < 4; i++) {
+            int nr = r + dr[i];
+            int nc = c + dc[i];
+            if (nr < 0 || nr >= m || nc < 0 || nc >= n || visited[nr][nc]) continue;
+            if (!curr.children.containsKey(board[nr][nc])) continue;
+
+            visited[nr][nc] = true;
+            sb.append(board[nr][nc]);
+            dfs(m, n, board, visited, nr, nc, curr.children.get(board[nr][nc]), sb, ans);
+            sb.deleteCharAt(sb.length() - 1);
+            visited[nr][nc] = false;
+        }
+    }
+}