Problem

给定一个由字符和单词组成的m x n 2D棋盘，找出该单词在棋盘中出现的次数.

单词可以由连续相邻单元的字母构成，其中相邻单元水平或垂直相邻.

注意:同一个字母单元格不能使用多次.

输入格式

第一行包含两个空格分隔的整数m，n，分别表示电路板中的行数和列数.

约束条件

1&lt；=1m、 n&lt；=n5

1&lt；=长度(单词)&lt；=30

enter code here

输出格式

在二维板上打印给定单词的总数.

Sample Input

3 4
HDST
ANEO
BENY
NEST  // word

Sample Output

2

我也不能满足测试用例

5 5
LACOO 
OOOCL
NLOOO
ECOCO
MECOO
COOLOOC

Output should be

28

My Code

 public static void main(String[] args) {
        int m, n;
        Scanner sc = new Scanner(System.in);
        m = sc.nextInt();
        n = sc.nextInt();

        char[][] board = new char[m][n + 1];
        for (int i = 0; i < m; i++) {
            String temp = sc.next();
            board[i] = temp.toCharArray();
        }

        String word = sc.next();

        System.out.print(new Result().countWord(board, word, m, n));
    }

static class Result {
        static boolean[][] visited;
        static int count = 0;

        int countWord(char board[][], String word, int m, int n) {
            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(searchWord(board, word, i, j, m, n, 0))
                            count++;
                    }
                }
            }
            return count;
        }

        static boolean searchWord(char board[][], String word, int i, int j, int m, int n, int index) {
            if (index == word.length()) {
                count++;
                return true;
            }

            if (i < 0 || j < 0 || i >= m || j >= n)
                return false;

            if (visited[i][j] || board[i][j] != word.charAt(index))
                return false;

            visited[i][j] = true;

            if (searchWord(board, word, i - 1, j, m, n, index + 1) ||
                    searchWord(board, word, i + 1, j, m, n, index + 1) ||
                    searchWord(board, word, i, j - 1, m, n, index + 1) ||
                    searchWord(board, word, i, j + 1, m, n, index + 1)) {
                return true;
            }

            visited[i][j] = false;
            return false;
        }
    }