286. 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 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.
   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

一刷
题解：以所有的gates为起始点做BFS

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        if(rooms.length == 0 || rooms[0].length == 0) return;
        Queue queue = new LinkedList<>();
        
        for(int i=0; i0 && rooms[row-1][col] == Integer.MAX_VALUE){
                rooms[row-1][col] = rooms[row][col] + 1;
                queue.add(new int[]{row-1, col});
            }
            if(row0 && rooms[row][col-1] == Integer.MAX_VALUE){
                rooms[row][col-1] = rooms[row][col] + 1;
                queue.add(new int[]{row, col-1});
            }
            if(col

二刷
从gates出发，BFS

class Solution {
    public void wallsAndGates(int[][] rooms) {
        int m = rooms.length;
        if(m == 0) return;
        int n = rooms[0].length;
        Queue queue = new LinkedList<>();
        for(int i=0; i=0 && rooms[row-1][col] == Integer.MAX_VALUE){
                rooms[row-1][col] = rooms[row][col] + 1;
                queue.add(new int[]{row-1, col});
            }
            if(row+1=0 && rooms[row][col-1] == Integer.MAX_VALUE){
                rooms[row][col-1] = rooms[row][col] + 1;
                queue.add(new int[]{row, col-1});
            }
            if(col+1

三刷

class Solution {
    class Cell{
        int x;
        int y;
        Cell(int x, int y){
            this.x = x;
            this.y = y;
        }
    }
    
    public void wallsAndGates(int[][] rooms) {
        if(rooms == null || rooms.length == 0 || rooms[0].length == 0) return;
        int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1}};
        Deque queue = new LinkedList<>();
        int m = rooms.length, n = rooms[0].length;
        for(int i=0; i=0 && x=0 && y