286. Walls and Gates

问题描述：

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.

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

 

解题思路：

这道题实际上是求最短路径，我们可以用BFS来解。（DFS也可以解）

用BFS的话就是使用队列来存储可更新点的位置。

为了更新距离，我们用一个计数器cnt来记录当前距离的点的个数，每访问一个，计数器减1。

对于访问的每一个点，我们要把可更新的（矩阵中数字大于当前距离的）压入队列并且更新下一个距离的点的个数计数器nextCnt

当计数器cnt为0时，说明当前距离的点全部访问更新完了，我们需要对距离自增1，并且将nextCnt 赋给 cnt 并且将其置0.

 

代码：

class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        if(rooms.empty() || rooms[0].empty())
            return;
        for(int i = 0; i < rooms.size(); i++){
            for(int j = 0; j < rooms[i].size(); j++){
                if(rooms[i][j] == 0)
                    bfs(rooms, i, j);
            }
        }
    }
private:
    void bfs(vector<vector<int>> &room, int i, int j){
        queue<pair<int,int>> q;
        q.push({i,j});
        int cnt = 1;
        int dis = 1;
        int nextCnt = 0;
        while(!q.empty()){
            pair<int,int> p = q.front();
            q.pop();
            cnt--;
            if(p.first - 1 >= 0 && room[p.first-1][p.second] > dis){
                room[p.first-1][p.second] = dis;
                q.push({p.first-1, p.second});
                nextCnt++;
            }
            if(p.second - 1 >= 0 && room[p.first][p.second-1] > dis){
                room[p.first][p.second-1] = dis;
                q.push({p.first, p.second-1});
                nextCnt++;
            }
            if(p.first + 1< room.size() && room[p.first+1][p.second] > dis){
                room[p.first+1][p.second] = dis;
                q.push({p.first+1, p.second});
                nextCnt++;
            }
            if(p.second + 1 < room[0].size() && room[p.first][p.second+1] > dis){
                room[p.first][p.second+1] = dis;
                q.push({p.first, p.second+1});
                nextCnt++;
            }
            if(cnt == 0){
                dis++;
                cnt = nextCnt;
                nextCnt = 0;
            }
        }
    }
};