Question Definition
Given a 2D board containing ‘X’ and ‘O’ (the letter O), capture all regions surrounded by ‘X’.
A region is captured by flipping all ‘O’s into ‘X’s in that surrounded region.
For example,
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:
X X X X
X X X X
X X X X
X O X X
Java Solution
public void solve(char[][] board) {
if (board.length == 0 || board[0].length == 0)
return;
if (board.length < 2 || board[0].length < 2)
return;
int m = board.length, n = board[0].length;
//Any 'O' connected to a boundary can't be turned to 'X', so ...
//Start from first and last column, turn 'O' to '*'.
for (int i = 0; i < m; i++) {
if (board[i][0] == 'O')
boundaryDFS(board, i, 0);
if (board[i][n-1] == 'O')
boundaryDFS(board, i, n-1);
}
//Start from first and last row, turn '0' to '*'
for (int j = 0; j < n; j++) {
if (board[0][j] == 'O')
boundaryDFS(board, 0, j);
if (board[m-1][j] == 'O')
boundaryDFS(board, m-1, j);
}
//post-prcessing, turn 'O' to 'X', '*' back to 'O', keep 'X' intact.
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == '*')
board[i][j] = 'O';
}
}
}
//Use DFS algo to turn internal however boundary-connected 'O' to '*';
private void boundaryDFS(char[][] board, int i, int j) {
if (i < 0 || i > board.length - 1 || j <0 || j > board[0].length - 1)
return;
if (board[i][j] == 'O')
board[i][j] = '*';
if (i > 1 && board[i-1][j] == 'O')
boundaryDFS(board, i-1, j);
if (i < board.length - 2 && board[i+1][j] == 'O')
boundaryDFS(board, i+1, j);
if (j > 1 && board[i][j-1] == 'O')
boundaryDFS(board, i, j-1);
if (j < board[i].length - 2 && board[i][j+1] == 'O' )
boundaryDFS(board, i, j+1);
}
Comments