[Leetcode 286]: Walls and Gates

Walls and Gates

Total Accepted: 411 Total Submissions: 1365 Difficulty: Medium

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
两种思路：1.深搜，记忆化搜索。 2. 广搜，每次从门处搜（https://segmentfault.com/a/1190000003906674）
const int INF=2<<31-1;
class Solution {
public:
int m,n;
void wallsAndGates(vector<vector<int>>& grid) {
m=grid.size();
if(m==0) return;
n=grid[0].size();
vector<vector<bool>> vis(m,vector<bool>(n,true));
for(int i=0;i<m;i++) {
for(int j=0;j<n;j++) {
if(vis[i][j]&&grid[i][j]==INF) {
dfs(i,j,grid,vis);
}
}
}
}

int dfs(int x,int y,vector<vector<int>>& grid,vector<vector<bool>>& vis) {
vis[x][y]=false;
if(grid[x][y]>=0) return grid[x][y]; //说明该点已计算过，直接返回结果
int u,d,l,r;
u=d=l=r=INF;
if(x-1>=0) {
if(vis[x-1][y]&&grid[x-1][y]!=-1)
u=dfs(x-1,y,grid,vis);
}
if(x+1<m) {
if(vis[x+1][y]&&grid[x+1][y]!=-1)
d=dfs(x+1,y,grid,vis);
}
if(y-1>=0) {
if(vis[x][y-1]&&grid[x][y-1]!=-1)
l=dfs(x,y-1,grid,vis);
}
if(y+1<n) {
if(vis[x][y+1]&&grid[x][y+1]!=-1)
r=dfs(x,y+1,grid,vis);
}
grid[x][y]=min(min(min(u,d),l),r);
if(grid[x][y]!=INF) grid[x][y]+=1;
}
};