[leetcode] 286. Walls and Gates 解题报告

题目链接：https://leetcode.com/problems/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

思路：依然是一个DFS，遍历数组中的每一个元素，碰到原始是０的，也就是门，就以这个点开始做DFS．然后向上下左右方向走，需要注意的是为了防止循环调用，我们需要判断一下如果下一个位置的元素比当前元素小或者相等，就没有必要再去那个位置了．

代码如下：

class Solution {
public:
    void DFS(vector<vector<int>>& rooms, int i, int j, int val)
    {
        if(i<0 || i>=rooms.size() || j<0 || j>=rooms[0].size() 
            || rooms[i][j] == -1 || rooms[i][j]<val)
                return;
        rooms[i][j] = min(rooms[i][j], val);
        DFS(rooms, i, j+1, rooms[i][j]+1);
        DFS(rooms, i, j-1, rooms[i][j]+1);
        DFS(rooms, i+1, j, rooms[i][j]+1);
        DFS(rooms, i-1, j, rooms[i][j]+1);
    }
    void wallsAndGates(vector<vector<int>>& rooms) {
        if(rooms.size()==0) return;
        for(int i =0; i< rooms.size(); i++)
            for(int j =0; j< rooms[0].size(); j++)
                if(rooms[i][j] == 0)
                    DFS(rooms, i, j, 0);
    }
};