leetcode wordSearch

public class wordSearch {
    /**
     * 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.
     *
     * Example:
     *
     * board =
     * [
     *   ['A','B','C','E'],
     *   ['S','F','C','S'],
     *   ['A','D','E','E']
     * ]
     *
     * Given word = "ABCCED", return true.
     * Given word = "SEE", return true.
     * Given word = "ABCB", return false.
     */
    public boolean exist(char[][] board, String word) {
        if (board == null || board[0].length == 0) return false;
        int m = board.length;
        int n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (board[i][j] == word.charAt(0)) {
                    if (dfs(word, board, visited, m, n, i, j, 0)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }

    public boolean dfs(String word, char[][] board, boolean[][] visited, int m, int n, int i, int j, int index) {
        if (i < 0 || j < 0 || i >= m || j>=n || visited[i][j] || board[i][j]!=word.charAt(index))
            return false;
        if (index == word.length()-1) return true;
        visited[i][j] = true;

        boolean ans =  dfs(word, board, visited, m, n, i+1, j, index+1)
                || dfs(word, board, visited, m, n, i-1, j, index+1)
                || dfs(word, board, visited, m, n, i, j+1, index+1)
                || dfs(word, board, visited, m, n, i, j-1, index+1) ;
        visited[i][j] = false;
        return ans;
    }
}