Word Search II

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example:

Input: 
words = ["oath","pea","eat","rain"] and board =
[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Output: ["eat","oath"]

Note: You may assume that all inputs are consist of lowercase lettersa-z.

Solution

class Solution {
    public List<String> findWords(char[][] board, String[] words) {
        List<String> res = new ArrayList<>();
        TrieNode root = buildTrie(words);
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                dfs (board, i, j, root, res);
            }
        }
        return res;
    }

    public void dfs(char[][] board, int i, int j, TrieNode p, List<String> res) {
        char c = board[i][j];
        if (c == '#' || p.next[c - 'a'] == null) return;
        p = p.next[c - 'a'];
        if (p.word != null) {   // found one
            res.add(p.word);
            p.word = null;     // de-duplicate
        }

        board[i][j] = '#';
        if (i > 0) dfs(board, i - 1, j ,p, res); 
        if (j > 0) dfs(board, i, j - 1, p, res);
        if (i < board.length - 1) dfs(board, i + 1, j, p, res); 
        if (j < board[0].length - 1) dfs(board, i, j + 1, p, res); 
        board[i][j] = c;
    }

    public TrieNode buildTrie(String[] words) {
        TrieNode root = new TrieNode();
        for (String w : words) {
            TrieNode p = root;
            for (char c : w.toCharArray()) {
                int i = c - 'a';
                if (p.next[i] == null) p.next[i] = new TrieNode();
                p = p.next[i];
           }
           p.word = w;
        }
        return root;
    }

    class TrieNode {
        TrieNode[] next = new TrieNode[26];
        String word;
    }
}