zoukankan      html  css  js  c++  java
  • 490. The Maze 迷宫踢足球

    There is a ball in a maze with empty spaces (represented as 0) and walls (represented as 1). The ball can go through the empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

    Given the m x n maze, the ball's start position and the destination, where start = [startrow, startcol] and destination = [destinationrow, destinationcol], return true if the ball can stop at the destination, otherwise return false.

    You may assume that the borders of the maze are all walls (see examples).

     

    Example 1:

    Input: maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [4,4]
    Output: true
    Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.
    

    Example 2:

    Input: maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [3,2]
    Output: false
    Explanation: There is no way for the ball to stop at the destination. Notice that you can pass through the destination but you cannot stop there.
    

    Example 3:

    Input: maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], start = [4,3], destination = [0,1]
    Output: false

    忘记这种dfs怎么写了。一句话:边界感!

    https://leetcode.com/problems/the-maze/discuss/132857/Java-DFS-solution-beats-98

    int[][] steps = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        
        public boolean hasPath(int[][] maze, int[] start, int[] destination) {
            //DFS + cache
            int rows = maze.length;
            int cols = maze[0].length;
            boolean[][] visited = new boolean[rows][cols];
            
            return dfs(maze, start, destination, visited, rows, cols);
        }
        
        private boolean dfs(int[][] maze, int[] start, int[] des, boolean[][] visited, int rows, int cols) {
            if (visited[start[0]][start[1]]) {
                return false;
            }
            if (start[0] == des[0] && start[1] == des[1]) {
                return true;
            }
            
            visited[start[0]][start[1]] = true;
            for (int[] step : steps) {
                int dx = start[0];
                int dy = start[1];
                while (dx + step[0] >= 0 && dx + step[0] < rows && dy + step[1] >= 0 && dy + step[1] < cols && maze[dx + step[0]][dy + step[1]] != 1) {
                    dx += step[0];
                    dy += step[1];
                } 
                if (dfs(maze, new int[]{dx, dy}, des, visited, rows, cols)) {
                    return true;
                }
            }
            return false;
        }



  • 相关阅读:
    bat批处理脚本学习系列(一)
    遇到的bug
    util.js 积累的一些基础函数代码
    session过期跳转到登陆页面并解决跳出iframe问题
    Centos7下的rabbitmq-server-3.8.11安装配置
    简单梳理 ES6 函数
    博客目录与学习计划
    DDD中聚合、聚合根的含义以及作用
    DDD中实体与值对象是干什么的
    DDD中限界上下文与通用语言的作用
  • 原文地址:https://www.cnblogs.com/immiao0319/p/15195979.html
Copyright © 2011-2022 走看看