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

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

解題思路：
用DFS或者BFS來解，思路是，我們搜索0的位置，每找到一個0，以其周圍四個相鄰點為起點，開始DFS遍歷，並帶入深度值1，如果遇到的值大於當前深度值，我們將位置值賦為當前深度值，並對當前點的四個相鄰點開始DFS遍歷，註意此時深度值需要加1，這樣遍歷完成後，所有的位置就被正確地更新了。

C++: DFS

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

C++: DFS II

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

C++: BFS

class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        queue<pair<int, int>> q;
        vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
        for (int i = 0; i < rooms.size(); ++i) {
            for (int j = 0; j < rooms[i].size(); ++j) {
                if (rooms[i][j] == 0) q.push({i, j});   
            }
        }
        while (!q.empty()) {
            int i = q.front().first, j = q.front().second; q.pop();
            for (int k = 0; k < dirs.size(); ++k) {
                int x = i + dirs[k][0], y = j + dirs[k][1];
                if (x < 0 || x >= rooms.size() || y < 0 || y >= rooms[0].size() || rooms[x][y] < rooms[i][j] + 1) continue;
                rooms[x][y] = rooms[i][j] + 1;
                q.push({x, y});
            }
        }
    }
};

[LeetCode] 286. Walls and Gates 墻和門