Last Updated on 2022-11-21 by Clay
題目
You are given an m x n
matrix maze
(0-indexed) with empty cells (represented as '.'
) and walls (represented as '+'
). You are also given the entrance
of the maze, where entrance = [entrancerow, entrancecol]
denotes the row and column of the cell you are initially standing at.
In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance
. An exit is defined as an empty cell that is at the border of the maze
. The entrance
does not count as an exit.
Return the number of steps in the shortest path from the entrance
to the nearest exit, or -1
if no such path exists.
Example 1:
Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2] Output: 1 Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3]. Initially, you are at the entrance cell [1,2]. - You can reach [1,0] by moving 2 steps left. - You can reach [0,2] by moving 1 step up. It is impossible to reach [2,3] from the entrance. Thus, the nearest exit is [0,2], which is 1 step away.
Example 2:
Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0] Output: 2 Explanation: There is 1 exit in this maze at [1,2]. [1,0] does not count as an exit since it is the entrance cell. Initially, you are at the entrance cell [1,0]. - You can reach [1,2] by moving 2 steps right. Thus, the nearest exit is [1,2], which is 2 steps away.
Example 3:
Input: maze = [[".","+"]], entrance = [0,0] Output: -1 Explanation: There are no exits in this maze.
Constraints:
maze.length == m
maze[i].length == n
1 <= m, n <= 100
maze[i][j]
is either'.'
or'+'
.entrance.length == 2
0 <= entrancerow < m
0 <= entrancecol < n
entrance
will always be an empty cell.
題目會給定一張迷宮、起始座標,然後讓你判斷最短抵達迷宮出口的步數是多少?
迷宮出口的定義就是接觸到迷宮的邊界、該邊界並非牆壁,牆壁是不能走的。另外,出發點不能是出口。
解題思路
迷宮探路的題目會很直覺地想到使用 BFS 來找到最近的出口,因為 BFS 是一層層地往外去探索可能性,找到的第一個出口也就會是最短的路線。
C++ 範例程式碼
class Solution {
public:
int nearestExit(vector<vector<char>>& maze, vector<int>& entrance) {
// Offset
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
// Init
int steps = 0;
int _x = entrance[0];
int _y = entrance[1];
int width = maze.size();
int height = maze[0].size();
queue<pair<int, int>> q;
q.push({_x, _y});
vector<vector<bool>> isVisited(width, vector<bool>(height, 0));
isVisited[_x][_y] = 1;
// BFS
while (!q.empty()) {
int size = q.size();
++steps;
while (size) {
--size;
_x = q.front().first;
_y = q.front().second;
q.pop();
for (int i=0; i<4; ++i) {
int new_x = _x + dx[i];
int new_y = _y + dy[i];
// Out of the boundary
if (new_x < 0 || new_y < 0 || new_x >= width || new_y >= height) {
continue;
}
// Wall or is visited
if (maze[new_x][new_y] == '+' || isVisited[new_x][new_y]) {
continue;
}
// Reach the exit
if (new_x == 0 || new_y == 0 || new_x == width-1 || new_y == height-1) {
return steps;
}
isVisited[new_x][new_y] = 1;
q.push({new_x, new_y});
}
}
}
return -1;
}
};
Python 範例程式碼
class Solution {
public:
int nearestExit(vector<vector<char>>& maze, vector<int>& entrance) {
// Offset
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
// Init
int steps = 0;
int _x = entrance[0];
int _y = entrance[1];
int width = maze.size();
int height = maze[0].size();
queue<pair<int, int>> q;
q.push({_x, _y});
vector<vector<bool>> isVisited(width, vector<bool>(height, 0));
isVisited[_x][_y] = 1;
// BFS
while (!q.empty()) {
int size = q.size();
++steps;
while (size) {
--size;
_x = q.front().first;
_y = q.front().second;
q.pop();
for (int i=0; i<4; ++i) {
int new_x = _x + dx[i];
int new_y = _y + dy[i];
// Out of the boundary
if (new_x < 0 || new_y < 0 || new_x >= width || new_y >= height) {
continue;
}
// Wall or is visited
if (maze[new_x][new_y] == '+' || isVisited[new_x][new_y]) {
continue;
}
// Reach the exit
if (new_x == 0 || new_y == 0 || new_x == width-1 || new_y == height-1) {
return steps;
}
isVisited[new_x][new_y] = 1;
q.push({new_x, new_y});
}
}
}
return -1;
}
};