原题：

/**
 * Created by pradhang on 7/4/2017.
 * Given a 2D board and a list of words from the dictionary, find all words in the board.
 * <p>
 * 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.
 * <p>
 * For example,
 * Given words = ["oath","pea","eat","rain"] and board =
 * <p>
 * [
 * ['o','a','a','n'],
 * ['e','t','a','e'],
 * ['i','h','k','r'],
 * ['i','f','l','v']
 * ]
 * Return ["eat","oath"].
 * Note:
 * You may assume that all inputs are consist of lowercase letters a-z.
 */

答案：

public class WordSearchII {
    private final int[] R = {0, 0, -1, 1};
    private final int[] C = {-1, 1, 0, 0};
    boolean[][] visited;
    private Set<String> dictionary;

    public static void main(String[] args) throws Exception {
        char[][] board = {{'o', 'a', 'a', 'n'}, {'e', 't', 'a', 'e'}, {'i', 'h', 'k', 'r'}, {'i', 'f', 'l', 'v'}};
        String[] words = {"oath", "pea", "eat", "rain"};
        System.out.println(new WordSearchII().findWords(board, words));
    }

    public List<String> findWords(char[][] board, String[] words) {
        dictionary = new HashSet<>();
        Trie trie = new Trie();
        for (String w : words) {
            trie.insert(w);
            dictionary.add(w);
        }
        visited = new boolean[board.length][board[0].length];
        Set<String> resultSet = new HashSet<>();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                dfs(i, j, board, resultSet, trie, String.valueOf(board[i][j]));
            }
        }
        return new ArrayList<>(resultSet);
    }

    private void dfs(int r, int c, char[][] board, Set<String> result, Trie trie, String s) {
        char newChar = board[r][c];
        Trie subTrie = trie.next(newChar);
        if (subTrie == null) return;
        visited[r][c] = true;
        if (dictionary.contains(s))
            result.add(s);
        for (int i = 0; i < 4; i++) {
            int newR = r + R[i];
            int newC = c + C[i];
            if (newR >= 0 && newC >= 0 && newR < board.length && newC < board[0].length) {
                if (!visited[newR][newC]) {
                    dfs(newR, newC, board, result, subTrie, s + board[newR][newC]);
                }
            }
        }
        visited[r][c] = false;
    }

    private class Trie {

        private Map<Character, Trie> map;

        /** * Initialize your data structure here. */
        public Trie() {
            map = new HashMap<>();
        }

        /** * Inserts a word into the trie. */
        public void insert(String word) {
            if (word != null) {
                add(0, word, word.length());
            }
        }

        private void add(int i, String word, int length) {
            if (i < length) {
                char c = word.charAt(i);
                Trie subTrie = map.get(c);
                if (subTrie == null) {
                    subTrie = new Trie();
                    map.put(c, subTrie);
                }
                subTrie.add(i + 1, word, length);
            } else map.put(null, new Trie()); //use null to indicate end of string
        }

        /** * Get next Trie node * * @param c char c * @return return Trie */
        public Trie next(char c) {
            return this.map.get(c);
        }
    }
}