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  -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：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 > 0 ? rooms[0].size() : 0;
        for (int i = 0; i < height; i++)
            for (int j = 0; j < width; j++)
                if (rooms[i][j] == 0) q.push(make_pair(i, j));
        int rowChange[] = {0, -1, 0, 1};
        int colChange[] = {1, 0, -1, 0};
        while (!q.empty()) {
            pair<int, int> cur = q.front();
            q.pop();
            int row = get<0>(cur), col = get<1>(cur);
            for (int i = 0; i < 4; i++) {
                int nextRow = row + rowChange[i];
                int nextCol = col + colChange[i];
                if (nextRow > -1 && nextRow < height &&
                    nextCol > -1 && nextCol < width && rooms[nextRow][nextCol] == inf) {
                        rooms[nextRow][nextCol] = rooms[row][col] + 1;
                        q.push(make_pair(nextRow, nextCol));
                    }
            }
        }
    }
};