# My solution by firstly searching the boundary ones

• This is my code. I start searching from the boundary and mark 'O's as 'C'. Then the second traverse make the 'O' to 'X' and 'C' to 'O'.

``````class Solution {
public:
void solve(vector<vector<char> > &board) {
int row = board.size();
if (row == 0) return;
int col = board.front().size();
queue<pair<int,int> > neighbs;
int i, j, iner_i, iner_j;
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
if (i ==0 || i == (row-1) || j == 0 || j == (col-1)) {
if (board[i][j] == 'O') {
neighbs.push(make_pair(i,j));
while (!neighbs.empty()) {
iner_i = neighbs.front().first;
iner_j = neighbs.front().second;
neighbs.pop();
if (iner_i < 0 || iner_i > (row-1) || iner_j < 0 || iner_j > (col-1) || board[iner_i][iner_j] != 'O')
continue;
board[iner_i][iner_j] = 'C';
neighbs.push(make_pair(iner_i-1,iner_j));
neighbs.push(make_pair(iner_i+1,iner_j));
neighbs.push(make_pair(iner_i,iner_j-1));
neighbs.push(make_pair(iner_i,iner_j+1));
}
}
}
}
}
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
if (board[i][j] == 'O') {
board[i][j] = 'X';
}
if (board[i][j] == 'C') {
board[i][j] = 'O';
}
}
}
}
};``````

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