LeetCode 题解(259) : Walls and Gates

题目：

ou 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 representINF 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 withINF.

For example, given the 2D grid:

INF  -1  0  INFINF INF INF  -1INF  -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

题解：

C++版：

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