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  INFINF INF INF  -1INF  -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

对的

这道题看到tag有点奇怪，明明可以用dfs的。于是用dfs做出来了。需要注意的是对于0，也就是门的情况，不必特意考虑。因为在递归的时候，带入的层数是从1开始的，也就意味着如果碰到了门，0<1不会更新，所以不会出现stackoverflow。这种类型的题，除了要在递归函数的前面声明边界终止条件，还需要搞清楚层与层之间是如何传递的。

代码：

 public void wallsAndGates(int[][] rooms) {        if(rooms == null || rooms.length == 0 || rooms[0].length == 0) return;        int row = rooms.length;        int col = rooms[0].length;        for(int i=0;i<row;i++){            for(int j=0;j<col;j++){                if(rooms[i][j] == 0){                    dfs(rooms, i, j+1, 1);                    dfs(rooms, i+1, j, 1);                    dfs(rooms, i-1, j, 1);                    dfs(rooms, i, j-1, 1);                }            }        }    }    private void dfs(int[][] rooms, int i, int j, int curValue){        if(i<0 || j<0 || i>=rooms.length || j>=rooms[0].length) return;        if(rooms[i][j] == -1) return;        if(rooms[i][j]>curValue){            rooms[i][j] = curValue;            dfs(rooms, i+1, j, curValue+1);            dfs(rooms, i, j+1, curValue+1);            dfs(rooms, i-1, j, curValue+1);            dfs(rooms, i, j-1, curValue+1);        }    }