单词搜索

79. 单词搜索

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

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 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

class Solution {
    int[][] move = {{1,0}, {-1,0}, {0,1}, {0,-1}};
    boolean[][] visited;
    public boolean exist(char[][] board, String word) {
        visited = new boolean[board.length][board[0].length];
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(dfs(board, word, i, j, 0)){
                    return true;
                }
            }
        }
        return false;
 
    }
    public boolean dfs(char[][] board, String word, int x, int y, int k){
        if(x < 0 || x >= board.length || y < 0 || y >= board[0].length || visited[x][y]){
            return false;
        }
        if(board[x][y] != word.charAt(k)){
            return false;
        }else if(k == word.length()-1){
            return true;
        }
        visited[x][y] = true;
        for(int i = 0; i < 4; i++){
            int next_x = x + move[i][0];
            int next_y = y + move[i][1];
            if(dfs(board, word, next_x, next_y, k+1)){
                return true;
            }
        }
        visited[x][y] = false;
        return false;
    }
}