lc-212

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words， 返回所有二维网格上的单词 。

单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

示例 1：

输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]

示例 2：

输入：board = [["a","b"],["c","d"]], words = ["abcb"]
输出：[]

提示：

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] 是一个小写英文字母
• 1 <= words.length <= 3 * 104
• 1 <= words[i].length <= 10
• words[i] 由小写英文字母组成
• words 中的所有字符串互不相同

class Solution {
    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    public List<String> findWords(char[][] board, String[] words) {
        Trie trie = new Trie();
        for (String word : words) {
            trie.insert(word);
        }

        Set<String> ans = new HashSet<String>();
        for (int i = 0; i < board.length; ++i) {
            for (int j = 0; j < board[0].length; ++j) {
                dfs(board, trie, i, j, ans);
            }
        }

        return new ArrayList<String>(ans);
    }

    public void dfs(char[][] board, Trie now, int i1, int j1, Set<String> ans) {
        if (!now.children.containsKey(board[i1][j1])) {
            return;
        }
        char ch = board[i1][j1];
        now = now.children.get(ch);
        if (!"".equals(now.word)) {
            ans.add(now.word);
        }

        board[i1][j1] = '#';
        for (int[] dir : dirs) {
            int i2 = i1 + dir[0], j2 = j1 + dir[1];
            if (i2 >= 0 && i2 < board.length && j2 >= 0 && j2 < board[0].length) {
                dfs(board, now, i2, j2, ans);
            }
        }
        board[i1][j1] = ch;
    }
}

class Trie {
    String word;
    Map<Character, Trie> children;
    boolean isWord;

    public Trie() {
        this.word = "";
        this.children = new HashMap<Character, Trie>();
    }

    public void insert(String word) {
        Trie cur = this;
        for (int i = 0; i < word.length(); ++i) {
            char c = word.charAt(i);
            if (!cur.children.containsKey(c)) {
                cur.children.put(c, new Trie());
            }
            cur = cur.children.get(c);
        }
        cur.word = word;
    }
}