算法 leetcode212 单词搜索 II

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，找出所有同时在二维网格和字典中出现的单词。

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。

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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"]

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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 findWords(char[][] board, String[] words) {
        //构造前缀树，也是字典树
        Trie trie = new Trie();
        for (String word : words) {
            trie.insert(word);
        }
        //结果集
        Set ans = new HashSet<>();
        for (int i = 0; i < board.length; ++i) {
            for (int j = 0; j < board[0].length; ++j) {
                //从每个点进行dfs
                dfs(board, trie, i, j, ans);
            }
        }
        return new ArrayList<>(ans);
    }

    public void dfs(char[][] board, Trie now, int i1, int j1, Set ans) {
        //若当前节点char不是前缀树的第一层children的值，直接减枝
        if (!now.children.containsKey(board[i1][j1])) {
            return;
        }
        char ch = board[i1][j1];
        //获取对应字符下的前缀树
       now = now.children.get(ch);
        if (!"".equals(now.word)) {
            ans.add(now.word);
        }

        //将刚读过的字符置为#标记
        board[i1][j1] = '#';
        //对board[i][j]字符 上下左右移动，对后面点 dfs
        for (int[] dir : dirs) {
            int i2 = i1 + dir[0], j2 = j1 + dir[1];
            if (i2 >=0 && i2 < board.length && j2 >=0 && j2 children;
        boolean isWord;

        public Trie() {
            this.word = "";
            this.children = new HashMap();
        }

        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;
        }
    }

}