LEETCODE#130 surrounded regions

Original title Address

1. Replace All "O" with the border with other characters, such as "#"

2. Replace all remaining "O" with "X"

3. Replace All "#" with the border with "X"

Code:

1 voidReplace (vector<vector<Char> > &board,intIintJCharBefore,CharAfter ) {2stack<pair<int,int> >St;3   intRlen =board.size ();4   intClen = board[0].size ();5   intrnext[4] = {-1,1,0,0};6   intcnext[4] = {0,0, -1,1};7 8St.push (pair<int,int>(i, J));9    while(!St.empty ()) {Tenpair<int,int> top =st.top (); One St.pop (); A     if(Top.first <0|| Top.first >= Rlen | | Top.second <0|| Top.second >= Clen | | Board[top.first][top.second]! =before) -       Continue; -Board[top.first][top.second] =After ; the      for(intK =0; K <4; k++) -St.push (pair<int,int> (Top.first + rnext[k], Top.second +cnext[k])); -   } - } +  - voidSolve (vector<vector<Char>> &Board) { +   if(Board.empty () | | board[0].empty ())return; A   intRlen =board.size (); at   intClen = board[0].size (); -  -    for(inti =0; i < Rlen; i++) { -Replace (board, I,0,'O','#'); -Replace (board, I, Clen-1,'O','#'); -   } in    for(intj =0; J < Clen; J + +) { -Replace (board,0J'O','#'); toReplace (board, Rlen-1J'O','#'); +   } -  the    for(inti =1; I < Rlen-1; i++) *      for(intj =1; J < Clen-1; J + +) $       if(Board[i][j] = ='O')Panax NotoginsengBOARD[I][J] ='X'; -  the    for(inti =0; i < Rlen; i++) { +Replace (board, I,0,'#','O'); AReplace (board, I, Clen-1,'#','O'); the   } +    for(intj =0; J < Clen; J + +) { -Replace (board,0J'#','O'); $Replace (board, Rlen-1J'#','O'); $   } -}

At first it was replaced with a recursive return, the result exploded stack (Runtime error), had to honestly use the stack.

LEETCODE#130 surrounded regions