zoukankan      html  css  js  c++  java
  • Leetcode: The Maze II

    There is a ball in a maze with empty spaces and walls. The ball can go through 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 ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1.
    
    The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.
    
    Example 1
    
    Input 1: a maze represented by a 2D array
    
    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
    
    Input 2: start coordinate (rowStart, colStart) = (0, 4)
    Input 3: destination coordinate (rowDest, colDest) = (4, 4)
    
    Output: 12
    Explanation: One shortest way is : left -> down -> left -> down -> right -> down -> right.
                 The total distance is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.
    
    Example
    2 Input 1: a maze represented by a 2D array 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 Input 2: start coordinate (rowStart, colStart) = (0, 4) Input 3: destination coordinate (rowDest, colDest) = (3, 2) Output: -1 Explanation: There is no way for the ball to stop at the destination.
    Note: There is only one ball and one destination in the maze. Both the ball and the destination exist on an empty space, and they will not be at the same position initially. The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls. The maze contains at least
    2 empty spaces, and both the width and height of the maze won't exceed 100.

    Solution of The Mazehttps://discuss.leetcode.com/topic/77471/easy-understanding-java-bfs-solution
    Solution of The Maze IIIhttps://discuss.leetcode.com/topic/77474/similar-to-the-maze-ii-easy-understanding-java-bfs-solution

    We need to use PriorityQueue instead of standard queue, and record the minimal length of each point.

     1 public class Solution {
     2     class Point {
     3         int x,y,l;
     4         public Point(int _x, int _y, int _l) {x=_x;y=_y;l=_l;}
     5     }
     6     public int shortestDistance(int[][] maze, int[] start, int[] destination) {
     7         int m=maze.length, n=maze[0].length;
     8         int[][] length=new int[m][n]; // record length
     9         for (int i=0;i<m*n;i++) length[i/n][i%n]=Integer.MAX_VALUE;
    10         int[][] dir=new int[][] {{-1,0},{0,1},{1,0},{0,-1}};
    11         PriorityQueue<Point> list=new PriorityQueue<>((o1,o2)->o1.l-o2.l); // using priority queue
    12         list.offer(new Point(start[0], start[1], 0));
    13         while (!list.isEmpty()) {
    14             Point p=list.poll();
    15             if (length[p.x][p.y]<=p.l) continue; // if we have already found a route shorter
    16             length[p.x][p.y]=p.l;
    17             for (int i=0;i<4;i++) {
    18                 int xx=p.x, yy=p.y, l=p.l;
    19                 while (xx>=0 && xx<m && yy>=0 && yy<n && maze[xx][yy]==0) {
    20                     xx+=dir[i][0];
    21                     yy+=dir[i][1];
    22                     l++;
    23                 }
    24                 xx-=dir[i][0];
    25                 yy-=dir[i][1];
    26                 l--;
    27                 list.offer(new Point(xx, yy, l));
    28             }
    29         }
    30         return length[destination[0]][destination[1]]==Integer.MAX_VALUE?-1:length[destination[0]][destination[1]];
    31     }
    32 }
  • 相关阅读:
    [Todo]很不错的Java面试题类型整理,要看
    [Todo] Java并发编程学习
    自建一个Java Spring MVC项目
    [Todo] 乐观悲观锁,自旋互斥锁等等
    [Todo] Redis里面队列的两种模式,以及抢红包在Redis中的实现
    hdu 4704 同余定理+普通快速幂
    [置顶] ubuntu 和 win7 远程登陆 + vnc登陆
    mysql之触发器
    Jsoup API解析HTML中input标签
    IOS UITableView单条刷新,数据不刷新解决方案
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6396045.html
Copyright © 2011-2022 走看看