LeetCode Q286: Walls and Gates

You are given a m x n 2D grid initialized with these three possible values.

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF 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 with INF.

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:
BFS

	class Solution {
	public:
	void Fill(vector<vector<int>>& rooms, int entry, int num, int width, int height){
	int row = entry/width;
	int col = entry%width;
	rooms[row][col] = num;
	if(row-1>=0 && rooms[row-1][col]>(num+1))
	Fill(rooms, (row-1)*width+col, num+1, width, height);
	if(col-1>=0 && rooms[row][col-1]>(num+1))
	Fill(rooms, row*width+col-1, num+1, width, height);
	if(row+1<=height-1 && rooms[row+1][col]>(num+1))
	Fill(rooms, (row+1)*width+col, num+1, width, height);
	if(col+1<=width-1 && rooms[row][col+1]>(num+1))
	Fill(rooms, row*width+col+1, num+1, width, height);
	}
	void wallsAndGates(vector<vector<int>>& rooms) {
	if(rooms.empty())
	return;
	int m = rooms.size();
	int n = rooms[0].size();
	for(int i=0; i<m; i++)
	for(int j=0; j<n; j++)
	if(rooms[i][j]==0)
	Fill(rooms, i*n+j, 0, n, m);
	}
	};

view raw Q286.cpp hosted with ❤ by GitHub