You are given a m x n 2D grid initialized with these three possible values.
- -1 - A wall or an obstacle.
- 0 - A gate.
- INF - Infinity means an empty room. We use the value 231 - 1 =
2147483647
to representINF
as you may assume that the distance to a gate is less than2147483647
. Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled withINF
.
For example, given the 2D grid:
INF -1 0 INF
INF INF INF -1
INF -1 INF -1
0 -1 INF INF
After running your function, the 2D grid should be:
3 -1 0 1
2 2 1 -1
1 -1 2 -1
0 -1 3 4
Solution C++
#include <iostream>
#include <vector>
#include <queue>
#include <utility>
using namespace std;
#define INF 100
void push_to_queue(vector<vector<int>>& rooms, queue<pair<int, int>>& q, int x, int y, int d){
if(x >= 0 && x < rooms.size() && y >= 0 && y < rooms[0].size() && rooms[x][y] > d){
q.push({x, y});
rooms[x][y] = d;
}
}
void bfs(vector<vector<int>>& rooms, int i, int j, int dist){
queue<pair<int, int>> gates;
gates.push({i, j});
while(!gates.empty()){
int count = gates.size();
for(int _x = 0; _x < count; ++_x){
pair<int, int> pos = gates.front();
gates.pop();
int x = pos.first;
int y = pos.second;
push_to_queue(rooms, gates, x, y+1, dist+1);
push_to_queue(rooms, gates, x, y-1, dist+1);
push_to_queue(rooms, gates, x+1, y, dist+1);
push_to_queue(rooms, gates, x-1, y, dist+1);
}
dist++;
}
}
void wallsAndGates(vector<vector<int>>& rooms) {
for(int i = 0; i < rooms.size(); ++i){
for(int j = 0; j < rooms[0].size(); ++j){
if(!rooms[i][j])
bfs(rooms, i, j, 0);
}
}
}
void print_rooms(vector<vector<int>> rooms){
for(int i = 0; i < rooms.size(); ++i){
for(int j = 0; j < rooms[0].size(); ++j){
cout << rooms[i][j] << "\t";
}
cout << "\n";
}
cout << "-------------------\n";
}
int main() {
vector<vector<int>> rooms = {
{INF, -1, 0, INF},
{INF, INF, INF, -1},
{INF, -1, INF, -1},
{0, -1, INF, INF}
};
print_rooms(rooms);
wallsAndGates(rooms);
print_rooms(rooms);
return 0;
}
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