1. Description

Given a 2D board and a word, find if the word exists in the grid.

The word can 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.

2. Example

Given board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

3. Code

public class LeetCode0079 {

    private final int[][] direction = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };

    public boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0 || board[0].length == 0 || word == null || word.length() == 0) {
            return false;
        }
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (dfs(board, i, j, word, 0)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean dfs(char[][] board, int i, int j, String word, int pos) {

        if (pos >= word.length()) {
            return true;
        }
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length) {
            return false;
        }
        if (word.charAt(pos) != board[i][j]) {
            return false;
        }
        board[i][j] = '#';
        for (int k = 0; k < direction.length; k++) {
            if (dfs(board, i + direction[k][0], j + direction[k][1], word, pos + 1)) {
                return true;
            }
        }
        board[i][j] = word.charAt(pos);
        return false;
    }

    public static void main(String[] args) {
        LeetCode0079 leetcode = new LeetCode0079();
        char[][] board = new char[][] { { 'A', 'B', 'C', 'E' }, { 'S', 'F', 'C', 'S' }, { 'A', 'D', 'E', 'E' } };
        System.out.println(leetcode.exist(board, "SEE"));
    }
}