[LeetCode] 499. The Maze III 迷宫 III
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up (u), down (d), left (l) or right (r), but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction. There is also a hole in this maze. The ball will drop into the hole if it rolls on to the hole.
Given the ball position, the hole position and the maze, find out how the ball could drop into the hole by moving the shortest distance. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the hole (included). Output the moving directions by using 'u', 'd', 'l' and 'r'. Since there could be several different shortest ways, you should output the lexicographically smallest way. If the ball cannot reach the hole, output "impossible".
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 ball and the hole coordinates are represented by row and column indexes.
Example 1
Input 1: a maze represented by a 2D array 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 Input 2: ball coordinate (rowBall, colBall) = (4, 3)
Input 3: hole coordinate (rowHole, colHole) = (0, 1) Output: "lul"
Explanation: There are two shortest ways for the ball to drop into the hole.
The first way is left -> up -> left, represented by "lul".
The second way is up -> left, represented by 'ul'.
Both ways have shortest distance 6, but the first way is lexicographically smaller because 'l' < 'u'. So the output is "lul".
Example 2
Input 1: a maze represented by a 2D array 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 Input 2: ball coordinate (rowBall, colBall) = (4, 3)
Input 3: hole coordinate (rowHole, colHole) = (3, 0)
Output: "impossible"
Explanation: The ball cannot reach the hole.
Note:
- There is only one ball and one hole in the maze.
- Both the ball and hole 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 the width and the height of the maze won't exceed 30.
490. The Maze 和 505. The Maze II 迷宫 II 的变形,在二维空间中放了洞,用u, r, d, l 这四个字母来分别表示上右下左,求让球掉入洞中的最小移动距离的移动方向字符串。在步数相等的情况下,返回按字母排序最小的答案。
解法1: BFS
解法2: DFS
Java:BFS,时间复杂度O(m * n * Max(m,n)),空间复杂度O(mn)
class Solution {
class Point {
int row, col, dist; public Point(int row, int col, int dist) {
this.row = row;
this.col = col;
this.dist = dist;
}
} public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
if (maze == null || maze.length == 0) {
return "";
} int m = maze.length, n = maze[0].length;
int[][] distance = new int[m][n]; for (int i = 0; i < m; i++) {
Arrays.fill(distance[i], Integer.MAX_VALUE);
} distance[ball[0]][ball[1]] = 0;
int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
String[] ways = {"u", "d", "l", "r"}; Map<Integer, String> map = new HashMap<>(); Queue<Point> queue = new LinkedList<>();
queue.add(new Point(ball[0], ball[1], 0)); findShortestWayByBFS(maze, queue, map, distance, hole, ways, directions); return map.containsKey(hole[0] * n + hole[1]) ? map.get(hole[0] * n + hole[1]) : "impossible";
} public void findShortestWayByBFS(int[][] maze, Queue<Point> queue, Map<Integer, String> map, int[][] distance, int[] hole, String[] ways, int[][] directions) {
int n = maze[0].length; while (!queue.isEmpty()) {
Point curPoint = queue.remove(); for (int i = 0; i < 4; i++) {
int row = curPoint.row, col = curPoint.col, dist = curPoint.dist;
String path = map.getOrDefault(row * n + col, ""); while (isValid(row, col, maze, hole)) {
row += directions[i][0];
col += directions[i][1];
++dist;
} if (row != hole[0] || col != hole[1]) {
row -= directions[i][0];
col -= directions[i][1];
--dist;
} path += ways[i]; if (dist < distance[row][col]) {
distance[row][col] = dist;
map.put(row * n + col, path);
queue.add(new Point(row, col, dist));
}
else if (dist == distance[row][col] && path.compareTo(map.getOrDefault(row * n + col, "")) < 0) {
map.put(row * n + col, path);
queue.add(new Point(row, col, dist));
}
}
}
} public boolean isValid(int row, int col, int[][] maze, int[] hole) {
return row >= 0 && row < maze.length && col >= 0 && col < maze[0].length && maze[row][col] == 0 && (row != hole[0] || col != hole[1]);
}
}
Java:
public class Solution {
public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
Queue<Point> queue = new LinkedList<>();
int row = maze.length;
int col = maze[0].length;
Point[][] points = new Point[row][col];
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
Point point = new Point(i, j);
points[i][j] = point;
if (i == hole[0] && j == hole[1]) {
point.path = "impossible";
}
}
}
int[] dirX = {0, -1, 1, 0}; // d, l, r, u
int[] dirY = {1, 0, 0, -1};
String[] dir = {"d", "l", "r", "u"};
Point startPoint = points[ball[0]][ball[1]];
startPoint.dis = 0;
queue.add(startPoint);
while (!queue.isEmpty()) {
Point point = queue.poll();
for (int i = 0; i < 4; i++) {
int x = point.x;
int y = point.y;
int dis = point.dis;
String path = point.path;
boolean inHole = false;
while (x >= 0 && x < row && y >= 0 && y < col && maze[x][y] == 0) {
x += dirX[i];
y += dirY[i];
dis++;
if (x == hole[0] && y == hole[1]) {
inHole = true;
break;
}
}
if (!inHole) {
x -= dirX[i];
y -= dirY[i];
dis--;
}
Point newPoint = points[x][y];
if (newPoint.dis > dis) {
newPoint.dis = dis;
newPoint.path = String.join("", point.path, dir[i]);
if (!inHole) {
queue.add(newPoint);
}
} else if (newPoint.dis == dis) {
boolean updated = false;
String newPath = String.join("", point.path, dir[i]);
for (int k = 0; k < newPoint.path.length() && k < newPath.length(); k++) {
if (newPoint.path.charAt(k) > newPath.charAt(k)) {
updated = true;
newPoint.path = newPath;
break;
}
}
if (!updated && newPoint.path.length() > newPath.length()) {
newPoint.path = newPath;
}
}
}
}
return points[hole[0]][hole[1]].path;
}
}
class Point {
int x;
int y;
String path;
int dis;
public Point(int x, int y) {
this.x = x;
this.y = y;
this.path = "";
this.dis = Integer.MAX_VALUE;
}
}
C++:
class Solution {
public:
string findShortestWay(vector<vector<int>>& maze, vector<int>& ball, vector<int>& hole) {
int m = maze.size(), n = maze[0].size();
vector<vector<int>> dists(m, vector<int>(n, INT_MAX));
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
vector<char> way{'l','u','r','d'};
queue<pair<int, int>> q;
unordered_map<int, string> u;
dists[ball[0]][ball[1]] = 0;
q.push({ball[0], ball[1]});
while (!q.empty()) {
auto t = q.front(); q.pop();
for (int i = 0; i < 4; ++i) {
int x = t.first, y = t.second, dist = dists[x][y];
string path = u[x * n + y];
while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == 0 && (x != hole[0] || y != hole[1])) {
x += dirs[i][0]; y += dirs[i][1]; ++dist;
}
if (x != hole[0] || y != hole[1]) {
x -= dirs[i][0]; y -= dirs[i][1]; --dist;
}
path.push_back(way[i]);
if (dists[x][y] > dist) {
dists[x][y] = dist;
u[x * n + y] = path;
if (x != hole[0] || y != hole[1]) q.push({x, y});
} else if (dists[x][y] == dist && u[x * n + y].compare(path) > 0) {
u[x * n + y] = path;
if (x != hole[0] || y != hole[1]) q.push({x, y});
}
}
}
string res = u[hole[0] * n + hole[1]];
return res.empty() ? "impossible" : res;
}
};
C++: DFS
class Solution {
public:
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
vector<char> way{'l','u','r','d'};
string findShortestWay(vector<vector<int>>& maze, vector<int>& ball, vector<int>& hole) {
int m = maze.size(), n = maze[0].size();
vector<vector<int>> dists(m, vector<int>(n, INT_MAX));
unordered_map<int, string> u;
dists[ball[0]][ball[1]] = 0;
helper(maze, ball[0], ball[1], hole, dists, u);
string res = u[hole[0] * n + hole[1]];
return res.empty() ? "impossible" : res;
}
void helper(vector<vector<int>>& maze, int i, int j, vector<int>& hole, vector<vector<int>>& dists, unordered_map<int, string>& u) {
if (i == hole[0] && j == hole[1]) return;
int m = maze.size(), n = maze[0].size();
for (int k = 0; k < 4; ++k) {
int x = i, y = j, dist = dists[x][y];
string path = u[x * n + y];
while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == 0 && (x != hole[0] || y != hole[1])) {
x += dirs[k][0]; y += dirs[k][1]; ++dist;
}
if (x != hole[0] || y != hole[1]) {
x -= dirs[k][0]; y -= dirs[k][1]; --dist;
}
path.push_back(way[k]);
if (dists[x][y] > dist) {
dists[x][y] = dist;
u[x * n + y] = path;
helper(maze, x, y, hole, dists, u);
} else if (dists[x][y] == dist && u[x * n + y].compare(path) > 0) {
u[x * n + y] = path;
helper(maze, x, y, hole, dists, u);
}
}
}
};
类似题目:
[LeetCode] 505. The Maze II 迷宫 II
All LeetCode Questions List 题目汇总
[LeetCode] 499. The Maze III 迷宫 III的更多相关文章
- [LeetCode] 505. The Maze II 迷宫 II
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- LeetCode 499. The Maze III
原题链接在这里:https://leetcode.com/problems/the-maze-iii/ 题目: There is a ball in a maze with empty spaces ...
- [LeetCode] 505. The Maze II 迷宫之二
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- 【LeetCode】556. Next Greater Element III 解题报告(Python)
[LeetCode]556. Next Greater Element III 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人 ...
- 3299: [USACO2011 Open]Corn Maze玉米迷宫
3299: [USACO2011 Open]Corn Maze玉米迷宫 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 137 Solved: 59[ ...
- [LeetCode] The Maze III 迷宫之三
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- LC 499. The Maze III 【lock,hard】
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- [LeetCode] The Maze II 迷宫之二
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- [LeetCode] 490. The Maze 迷宫
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
随机推荐
- 用Python添加写入数据到已经存在的Excel的xlsx文件
# coding:utf-8 from openpyxl import load_workbook import openpyxl # 写入已存在的xlsx文件第一种方法 # class Write_ ...
- 当电脑上有两个mysql时,如何让jmeter连接自己需要的那个mysql
1.当有两个mysql时,修改其中一个的mysql端口号为3307 ,也可以是其他8788, 在文件my.ini中修改端口号为:3307:修改完之后记得重启mysql哦:dos下命令可以用net st ...
- Json在序列化注意问题
Java中的Json序列化,不容忽视的getter 问题重现 public class AjaxJson { private boolean success; private String msg; ...
- java将图片输出base64位码显示
注意需要过滤:\r \n数据 jkd1.7的 import sun.misc.BASE64Decoder;import sun.misc.BASE64Encoder; /** * 网络图片转换Base ...
- JS全局变量是如何工作的?
JS全局变量是如何工作的? <script> const one = 1; var two = 2; </script> <script> // All scrip ...
- LSTM的结构
- 【CPLEX教程03】java调用cplex求解一个TSP问题模型
00 前言 前面我们已经搭建好cplex的java环境了,相信大家已经跃跃欲试,想动手写几个模型了.今天就来拿一个TSP的问题模型来给大家演示一下吧~ CPLEX系列教程可以关注我们的公众号哦!获取更 ...
- 基于python的学生管理系统(含数据库版本)
这次支持连接到后台的数据库,直接和数据库进行交互,实现基本的增删查改 #!/usr/bin/python3 # coding=utf-8 """ ************ ...
- 【洛谷】P5024 保卫王国 (倍增)
前言 传送门 很多人写了题解了,我就懒得写了,推荐一篇博客 那就分享一下我的理解吧(说得好像有人看一样 对于每个点都只有选与不选两种情况,所以直接用倍增预处理出来两种情况的子树之内,子树之外的最值,最 ...
- 第10组Alpha冲刺(1/4)
队名:凹凸曼 组长博客 作业博客 组员实践情况 童景霖 过去两天完成了哪些任务 文字/口头描述 学习Android studio和Java,基本了解APP前端的制作 完善项目APP原型 展示GitHu ...