Walls and Gates -- LeetCode

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  -    INF
INF INF INF  -
INF  - INF  -
    - INF INF

After running your function, the 2D grid should be:

    -
          -
    -     -
    -      

思路：BSF。

 class Solution {
 public:
     void wallsAndGates(vector<vector<int>>& rooms) {
         const int inf = INT_MAX;
         queue<pair<int, int> > q;
         int height = rooms.size();
         int width = height >  ? rooms[].size() : ;
         for (int i = ; i < height; i++)
             for (int j = ; j < width; j++)
                 if (rooms[i][j] == ) q.push(make_pair(i, j));
         int rowChange[] = {, -, , };
         int colChange[] = {, , -, };
         while (!q.empty()) {
             pair<int, int> cur = q.front();
             q.pop();
             int row = get<>(cur), col = get<>(cur);
             for (int i = ; i < ; i++) {
                 int nextRow = row + rowChange[i];
                 int nextCol = col + colChange[i];
                 if (nextRow > - && nextRow < height &&
                     nextCol > - && nextCol < width && rooms[nextRow][nextCol] == inf) {
                         rooms[nextRow][nextCol] = rooms[row][col] + ;
                         q.push(make_pair(nextRow, nextCol));
                     }
             }
         }
     }
 };