【leetcode-79】 单词搜索

（典型dfs）

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.

public class Solution {
    public boolean exist(char[][] board, String word) {
        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(dfs(board, word, visited, i, j, m, n, 0))
                   return true;
            }
        }
        return false;
    }
     
    private boolean dfs(char[][] board, String word, boolean[][] visited, int i, int j, int m, int n, int count){
        if(count == word.length()){
            return true;
        }
        if(i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word.charAt(count))
            return false;
        if(visited[i][j])
            return false;
        visited[i][j] = true;
        boolean res = dfs(board, word, visited, i - 1, j, m, n, count + 1) || 
            dfs(board, word, visited, i + 1, j, m, n, count + 1) || 
            dfs(board, word, visited, i, j - 1, m, n, count + 1)||
            dfs(board, word, visited, i, j + 1, m, n, count + 1);
        visited[i][j] = false;
        return res;
    }
}