286. Walls and Gates

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286. Walls and Gates

Description

You are given an m x n grid roomsinitialized with these three possible values.

  • -1A wall or an obstacle.
  • 0 A gate.
  • INF Infinity means an empty room. We use the value 2^31 - 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.

Example 1:

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Input: rooms = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1],[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]
Output: [[3,-1,0,1],[2,2,1,-1],[1,-1,2,-1],[0,-1,3,4]]

Example 2:

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Input: rooms = [[-1]]
Output: [[-1]]

Constraints:

  • m == rooms.length
  • n == rooms[i].length
  • 1 <= m, n <= 250
  • rooms[i][j] is -1, 0, or 2^31 - 1.

Hints/Notes

  • 2025/01/05
  • bfs
  • premium

Solution

Language: C++

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class Solution {
public:
    static constexpr int DIRs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

    void wallsAndGates(vector<vector<int>>& rooms) {
        int m = rooms.size(), n = rooms[0].size();
        queue<pair<int, int>> q;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (rooms[i][j] == 0) {
                    q.push({i, j});
                }
            }
        }
        int cur = 1;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                auto [x, y] = q.front();
                q.pop();
                for (int j = 0; j < 4; j++) {
                    int dx = x + DIRs[j][0], dy = y + DIRs[j][1];
                    if (dx >= 0 && dx < m && dy >= 0 && dy < n && rooms[dx][dy] == INT_MAX) {
                        q.push({dx, dy});
                        rooms[dx][dy] = cur;
                    }
                }
            }
            cur++;
        }
    }
};