題目描述:

請設計一個函式，用來判斷在一個矩陣中是否存在一條包含某字串所有字元的路徑。路徑可以從矩陣中的任意一格開始，每一步可以在矩陣中向左、右、上、下移動一格。如果一條路徑經過了矩陣的某一格，那麼該路徑不能再次進入該格子。

例如，在下面的3×4的矩陣中包含一條字串“bfce”的路徑（路徑中的字母用加粗標出）。

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

但矩陣中不包含字串“abfb”的路徑，因為字串的第一個字元b佔據了矩陣中的第一行第二個格子之後，路徑不能再次進入這個格子。

示例 1：

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

class Solution {

    public static void main(String[] args) {
        Solution solution = new Solution();
        char[][] board = {{'C','A','A'},{'A','A','A'},{'B','C','D'}};
        System.out.println(solution.exist(board,
 
"AAB"));
    }


    public boolean exist(char[][] board, String word) {
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                char c = board[i][j];
                if (c == word.charAt(0)){
                    boolean[][] visited = new
 
 boolean[board.length][board[0].length];
                    boolean  out = search(board,visited,word,0,i,j);
                    if (out){
                        return true;
                    }
                }
            }
        }
        boolean[][] visited = new boolean[board.length][board[0].length];
        boolean  out = search(board,visited,word,0,1,1);
        return false;
    }

    private boolean search(char[][] board, boolean[][] visited, String string, int index, int i, int j) {
        if (string.length() - 1== index){
            return true;
        }
        visited[i][j] = true;
        char c = string.charAt(index+1);
        int upi = i - 1>=0?(i-1):i;
        int dni = i + 1<board.length?(i + 1):i;
        int ltj = j - 1>=0?(j-1):j;
        int rtj = j + 1<board[0].length?(j + 1):j;

        if (!visited[upi][j] && board[upi][j] == c){
            boolean out = search(board, visited, string, index+1, upi, j);
            if (out) return true;
        }
        if (!visited[dni][j] && board[dni][j] == c){
            boolean out =  search(board, visited, string, index+1, dni, j);
            if (out) return true;
        }
        if (!visited[i][ltj] && board[i][ltj] == c){
            boolean out = search(board, visited, string, index+1, i, ltj);
            if (out) return true;
        }
        if (!visited[i][rtj] && board[i][rtj] == c){
            boolean out = search(board, visited, string, index+1, i, rtj);
            if (out) return true;
        }
        visited[i][j] = false;
        return false;
    }
}