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 represent INF as you may assume that the distance to a gate is less than 2147483647.
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.
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
Do a breath-first search.
class Solution(object): def wallsAndGates(self, rooms): """ :type rooms: List[List[int]] :rtype: void Do not return anything, modify rooms in-place instead. """ if len(rooms) == 0 or len(rooms) == 0: return currIndexes = list() empty = set() x, y = len(rooms), len(rooms) for i in range(x): for j in range(y): if rooms[i][j] == 2147483647: empty.add((i, j)) elif rooms[i][j] == 0: currIndexes.append((i, j)) dist = 1 while len(empty) > 0: if len(currIndexes) == 0: break curr = list() for i, j in currIndexes: self.process(i + 1, j, rooms, dist, empty, curr) self.process(i - 1, j, rooms, dist, empty, curr) self.process(i, j + 1, rooms, dist, empty, curr) self.process(i, j - 1, rooms, dist, empty, curr) dist += 1 currIndexes = curr def process(self, i, j, rooms, dist, empty, curr): if (i, j) in empty: rooms[i][j] = dist empty.remove((i, j)) curr.append((i, j))