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 than2147483647.

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

A typical topological sort question. Interesting.

There are some tricky corners we should keep in mind.

First, always test q.empty() before you want to access a element in queue.

Second, pair<int, int> is not hashable, so you should store map<int, int> and write your simple hash function (i * rooms[0].size() + j in my case)

Third, use an index to indicate the last node of current level in the queue. It is like algorithm in level order traversal of a tree.

1. Start from gates, push gates in queue, Set distance as 0. Set lastNodeinLevel to q.back()
2. while q is not empty:
   1. push legal neighbors of q.front() in queue and hash map. legal is defined as follows:
      1. is not out of boundary
      2. is not a wall (-1)
      3. does not ever existed in queue.
   2. set map[q.front().x][q.front().y] to distance
   3. if q.front() equal to lastNodeinLevel:
      1. distance++
      2. lastNodeinLevel = q.back() (if q is not empty yet)
   4. q.pop()

//topological sort
class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        //find all gates, and push the neighbors of gates in queue.(1. it's not in queue before, 2. it's not a -1)
        if(rooms.size() == 0 || rooms[0].size() == 0) return ;
        unordered_map<int, int> map;
        queue<pair<int, int>> q;
        pair<int, int> lastNodeinLevel;
        int distance = 0;
        for(int i = 0; i < rooms.size(); i++){
            for(int j = 0; j < rooms[0].size(); j++){
                if(rooms[i][j] == 0){
                    pair<int, int> nn = pair<int, int>(i, j);
                    q.push(nn);
                    map[i * rooms[0].size() + j] = 1;
                }
            }
        }
        if(q.empty()){//BUG HERE
           return ; 
        }
        lastNodeinLevel = q.back();
        distance = 0;
        while(!q.empty()){
            pair<int, int> node = q.front();
            q.pop();
            rooms[node.first][node.second] = distance;
            pushNeighborsInQueue(rooms, map, q, node);
            if(node == lastNodeinLevel && !q.empty()){
                lastNodeinLevel = q.back();
                distance++;
            }
        }
    }
    void pushNeighborsInQueue(vector<vector<int>>& rooms, unordered_map<int, int>& map, queue<pair<int, int>>& q, pair<int, int>& node){
        int direct[][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
        for(int i = 0; i < 4; i++){
            int x = node.first + direct[i][0];
            int y = node.second + direct[i][1];
            if(x < 0 || y < 0 || x >= rooms.size() || y >= rooms[0].size()){
                //out of boundary
                continue;
            }
            pair<int, int> nn = pair<int, int>(x, y);
            if(map.find(x * rooms[0].size() + y) != map.end()){
                continue;
            }
            if(rooms[x][y] == -1){
                continue;
            }
            q.push(nn);
            map[x * rooms[0].size() + y] = 1;
        }
        //already in map
        //is -1
    }
};