12. 矩阵中的路径

面试题12. 矩阵中的路径

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

提示：

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

class Solution {
    public boolean exist(char[][] board, String word) {
        if(word == null || word.length() == 0) return true;
        // List<Character> list = new ArrayList<>();
        boolean[][] 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,i,j,0,word,visited))
                    return true;
            }
        }
        return false;
        
    }
    
    private boolean dfs(char[][] board,int i, int j,int index,String word,boolean[][] visited){
        if(index == word.length()) return true;
        if(i >= board.length || i < 0 || j < 0 || j >= board[0].length || visited[i][j] || 
                   index >= word.length() || board[i][j] != word.charAt(index)){
            return false;
        }
        visited[i][j] = true;
        // list.add(board[i][j]);
        if(dfs(board,i + 1,j,index + 1,word,visited) || dfs(board,i,j + 1,index + 1,word,visited) ||
           dfs(board,i - 1,j,index + 1,word,visited) ||dfs(board,i,j - 1,index + 1,word,visited) ){
            return true;
        }
        // list.remove(list.size() - 1);
        visited[i][j] = false;
        return false;
    }
}