回溯法
打个卡

class Solution {
    private int[] dx = {1, -1, 0, 0};
    private int[] dy = {0, 0, 1, -1};
    private boolean[][] visited;
    public boolean hasPath(char[][] matrix, String str) {
        if(str.length() == 0)return true;
        if(matrix.length == 0 || matrix[0].length == 0)return false;
        int r = matrix.length, c = matrix[0].length;
        visited = new boolean[r][c];
        for(int i = 0; i < r; i ++)
            for(int j = 0; j < c; j ++)
                visited[i][j] = false;

        for(int i = 0; i < r; i ++)
            for(int j = 0; j < c; j ++){
                if(matrix[i][j] == str.charAt(0)){
                    visited[i][j] = true;
                    boolean res = dfs(matrix, str, 1, i, j);
                    visited[i][j] = false;
                    if(res)return true;
                }
            }

        return false;
    }

    private boolean dfs(char[][] matrix, String str, int curPos, int x, int y){
        if(curPos == str.length())return true;
        int r = matrix.length, c = matrix[0].length;
        boolean res = false;
        for(int i = 0; i < 4; i ++){
            int a = x + dx[i], b = y + dy[i];
            if(a >= 0 && a < r && b >= 0 && b < c && !visited[a][b] && matrix[a][b] == str.charAt(curPos)){
                visited[a][b] = true;
                res = res || dfs(matrix, str, curPos + 1, a, b);
                visited[a][b] = false;
            }
        }

        return res;
    }
}