一、题目

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

二、代码

class Solution {

    private int rowl, coll;
    private boolean[][] visited;
    private int[][] direction;

    public boolean exist(char[][] board, String word) {
        if(board == null || word == null || board.length == 0 || board[0].length == 0){
            return false;
        }
        rowl = board.length;
        coll = board[0].length;
        visited = new boolean[rowl][coll];
        direction = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        char[] mywords = word.toCharArray();
        for(int i = 0; i < rowl; i++){
            for(int j = 0; j < coll; j++){
                visited[i][j] = false;
            }
        }
        for(int row = 0; row < rowl; row++){
            for(int col = 0; col < coll; col++){
                boolean res = dfs(board, mywords, row, col, 0);
                if(res){
                    return true;
                }
            }
        }
        return false;
    }

    public boolean dfs(char[][] board, char[] mywords, int row, int col, int index){
        if(visited[row][col] || board[row][col] != mywords[index]){
            return false;
        }
        if(index == mywords.length - 1){
            return true;
        }
        visited[row][col] = true;
        boolean result = false;
        for(int[] dir : direction){
            int newr = row + dir[0];
            int newc = col + dir[1];
            if(newr >= 0 && newr < rowl && newc >= 0 && newc <coll && !visited[newr][newc]){
                boolean res2 = dfs(board, mywords, newr, newc, index+1);
                if(res2){
                    return true;
                }
            }
        }
        visited[row][col] = false;
        return false;
    }
}