130. Surrounded Regions

Given a 2D board containing 

'X'

 and 

'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

Solution 1 DFS
public void solve(char[][] board) {
if (board
4000
.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);
}



Solution 2 BFS
// BFS
static class Point {
int x;
int y;

Point(int x, int y) {
this.x = x;
this.y = y;
}
}

public static void solve2(char[][] board) {
if (board == null || board.length == 0)
return;
int rows = board.length, columns = board[0].length;
int[][] direction = { { -1, 0 }, { 1, 0 }, { 0, 1 }, { 0, -1 } };
for (int i = 0; i < rows; i++)
for (int j = 0; j < columns; j++) {
if ((i == 0 || i == rows - 1 || j == 0 || j == columns - 1) && board[i][j] == 'O') {
Queue<Point> queue = new LinkedList<>();
board[i][j] = 'B';
queue.offer(new Point(i, j));
while (!queue.isEmpty()) {
Point point = queue.poll();
for (int k = 0; k < 4; k++) {
int x = direction[k][0] + point.x;
int y = direction[k][1] + point.y;
if (x >= 0 && x < rows && y >= 0 && y < columns && board[x][y] == 'O') {
board[x][y] = 'B';
queue.offer(new Point(x, y));
}
}
}
}
}
for (int i = 0; i < rows; i++)
for (int j = 0; j < columns; j++) {
if (board[i][j] == 'B')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}

}