[Leetcode 286]: Walls and Gates

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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  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
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;}};

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