剑指 Offer 12. 矩阵中的路径(dfs)

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

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","d"]], word = "abcd"
输出：false

//v1.0
class Solution {
    int t = 0;
    int r,c;
    boolean[][] visited;
    char[][] board;
    public boolean exist(char[][] board, String word) {
        r = board.length;
        c = board[0].length;
        visited = new boolean[r][c];
        this.board = board;
        for(int i = 0; i < r; i++){
            for(int j = 0; j < c; j++){
                if(board[i][j] == word.charAt(0)){
                    // if(word.length() == 1) return true;
                    if(dfs(i,j,word)){
                        return true;
                    } 
                    // else{
                    //     visited = new boolean[r][c];
                    // }
                }
            }
        }
        return false;
    }

    public boolean dfs(int i, int j, String word){
        if(i < 0 || i >= r || j < 0 || j >= c || visited[i][j] || board[i][j] != word.charAt(t)) return false; 
        if(t == word.length() - 1) return true;
        // if(board[i][j] == word.charAt(t)){
            t++;
            visited[i][j] = true;      
            // return dfs(i+1,j,word) || dfs(i,j+1,word) || dfs(i-1,j,word) || dfs(i,j-1,word);
            boolean ans = dfs(i+1,j,word) || dfs(i,j+1,word) || dfs(i-1,j,word) || dfs(i,j-1,word);
            visited[i][j] = false;//******
            t--;//******
            return ans;
        // }
        // return t == word.length();
    }
}