[Locked] Walls and Gates

Walls and Gates 

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

1. -1-A wall or an obstacle.
2. 0-Gate A.
3. INF-Infinity means a empty room. We use the value of represent as assume that the 231 - 1 = 2147483647 distance to INF a gate are less than 2147483647 .

Fill each empty and the distance to its nearest gate. If It is the impossible to reach a gate, it should are 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 is:

  3  -1   0   1  2   2   1  -1  1  -1   2  -1  0  -1   3   4

Analysis:

Deep search Dfs, attention pruning, note overflow

Code:

//Deep Search, three cases return: the room number has been smaller than the current distance, walls and doors, accessed, and these three cases can be expressed in a formula Grids[i][j] < distvoidDFS (vector<vector<int> > &grids,intDistintIintj) {if(Grids[i][j] <Dist)return; GRIDS[I][J]=Dist; DFS (grids, dist+1, I, J +1); DFS (grids, dist+1, i +1, J); DFS (grids, dist+1, I, J-1); DFS (grids, dist+1I1, J); return;}voidDistancefromgate (vector<vector<int> > &grids) {    if(Grids.empty ())return; //set up a border sentryGrids.insert (Grids.begin (), vector<int> (grids[0].size (),-1)); Grids.push_back (Vector<int> (grids[0].size (),-1));  for(Auto &row:grids) {Row.insert (Row.begin (),-1); Row.push_back (-1); }    //recursion starting from every 0     for(inti =0; I < grids.size (); i++)         for(intj =0; J < grids[0].size (); J + +)            if(Grids[i][j] = =0) DFS (grids,0, I, J); //Remove Border Sentrygrids.erase (Grids.begin ());    Grids.pop_back ();  for(Auto &row:grids)        {Row.erase (Row.begin ());    Row.pop_back (); }    return;}

[Locked] Walls and Gates