单词搜索

79. 单词搜索

给定一个二维网格和一个单词,找出该单词是否存在于网格中。

单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

 

示例:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

给定 word = "ABCCED", 返回 true
给定 word = "SEE", 返回 true
给定 word = "ABCB", 返回 false

 

提示:

board 和 word 中只包含大写和小写英文字母。

1 <= board.length <= 200

1 <= board[i].length <= 200

1 <= word.length <= 10^3

public class T79 {
    public boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return word == null || word.length() == 0;
        }
        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)  && canreach(board, visited, i, j, 0, word)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean canreach(char[][] board, boolean[][] visited,
                             int i, int j, int matchLen, String word) {
        if (matchLen == word.length()) {
            return true;
        }
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length ||
                visited[i][j] || !(board[i][j] == word.charAt(matchLen))) {
            return false;
        }

        visited[i][j] = true;
        if (canreach(board, visited, i + 1, j, matchLen + 1, word) ||
            canreach(board, visited, i - 1, j, matchLen + 1, word) ||
            canreach(board, visited, i, j + 1, matchLen + 1, word) ||
            canreach(board, visited, i, j - 1, matchLen + 1, word)){
            return true;
        }
        visited[i][j] = false;
        return false;

    }

    public static void main(String[] args) {
        char[][] board = {{'C','A','A'},{'A','A','A'},{'B','C','D'}};
        String word = "AAB";
        T79 t79 = new T79();
        t79.exist(board, word);
    }
}

 

posted @ 2020-03-19 17:46  海绵爱上星  阅读(223)  评论(0编辑  收藏  举报