LeetCode 286. Walls and Gates

You are given a m x n 2D grid initialized with these three possible values.

-1 - A wall or an obstacle.
0 - A gate.
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：
1. 对二维的规则结构，用dfs遍历。

void wallsAndGates(vector<vector<int>>& rooms) {    //dfs    int m=rooms.size();    if(m==0) return;    vector<vector<int>> dir={{-1,0},{1,0},{0,-1},{0,1}};//注意区别{}和（）的使用，{16,2,77,29}或second (4,100)    for(int i=0;i<m;i++){        for(int j=0;j<n;j++){            if(rooms[i][j]==0)                dfs(rooms,i,j,m,n,dir,0);        }    }}void dfs(vector<vector<int>>& rooms, int i,int j, int m,int n,vector<vector<int>>&dir,int depth){    if(i>=m||i<0||j>=n||j<0||rooms[i][j]<depth) return;    rooms[i][j]=depth;    for(auto&d:dir){        dfs(rooms,i+d[0],j+d[1],m,n,dir,depth+1);    }}