286. Walls and Gates

JIAKAOBO

LeetCode

感谢赞助!

• ㊗️
• 大家
• offer
• 多多！

Problem

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.

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

Code

class Solution {
    public void wallsAndGates(int[][] rooms) {
        Queue<int[]> queue = new LinkedList<>();
        
        for(int i = 0; i < rooms.length; i++){
            for(int j = 0; j < rooms[0].length; j++){
                if(rooms[i][j] == 0){
                    queue.add(new int[]{i, j});
                }
            }
        }
        
        int[][] dirs = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
        
        int curr = 0;
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i = 0; i < size; i++){
                int[] loc = queue.poll();
                int row = loc[0];
                int col = loc[1];
                rooms[row][col] = Math.min(curr, rooms[row][col]);

                for(int[] dir : dirs){
                    int x = row + dir[0];
                    int y = col + dir[1];

                    if(x < 0 || y < 0 || x >= rooms.length || y >= rooms[0].length) continue;
                    if(rooms[x][y] != Integer.MAX_VALUE) continue;

                    queue.offer(new int[]{x, y});
                }
            }
            
            curr++;
        }
    }
}

class Solution:
    def helper(self, i: int, j: int, step: int, rooms: List[List[int]]) -> None:
        if i < 0 or i >= len(rooms) or j < 0 or j >= len(rooms[0]):
            return

        if rooms[i][j] < step:
            return

        rooms[i][j] = step

        self.helper(i + 1, j, step + 1, rooms)
        self.helper(i - 1, j, step + 1, rooms)
        self.helper(i, j + 1, step + 1, rooms)
        self.helper(i, j - 1, step + 1, rooms)


    def wallsAndGates(self, rooms: List[List[int]]) -> None:
        """
        Do not return anything, modify rooms in-place instead.
        """
        for i in range(len(rooms)):
            for j in range(len(rooms[0])):
                if rooms[i][j] == 0:
                    self.helper(i, j, 0, rooms)

class Solution:
    def wallsAndGates(self, rooms: List[List[int]]) -> None:
        """
        Do not return anything, modify rooms in-place instead.
        """
        queue = deque()

        for i in range(len(rooms)):
            for j in range(len(rooms[0])):
                if rooms[i][j] == 0:
                    queue.append((i, j))
        step = 0
        moves = [[1, 0], [-1, 0], [0, 1], [0, -1]]
        while len(queue):
            size = len(queue)
            for i in range(size):
                node = queue.popleft()
                for move_index in range(4):
                    move = moves[move_index]
                    next_i = node[0] + move[0]
                    next_j = node[1] + move[1]
                    if next_i < 0 or next_i >= len(rooms) or next_j < 0 or next_j >= len(rooms[0]):
                        continue

                    if rooms[next_i][next_j] < step + 1:
                        continue

                    rooms[next_i][next_j] = step + 1
                    queue.append((next_i, next_j))

            step += 1