Java for LeetCode 130 Surrounded Regions

標籤：

 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
解題思路：

對最外面一圈進行BFS，替換掉最外圈的連通範圍，剩下的‘O’都是被包圍的，就可以直接kill掉，JAVA實現如下：

static public void solve(char[][] board) {if (board.length <= 2 || board[0].length <= 2)return;for (int row = 0; row < board.length; row++) {if (board[row][0] == ‘O‘) {board[row][0] = ‘T‘;bfs(board, row * board[0].length);}if (board[row][board[0].length - 1] == ‘O‘) {board[row][board[0].length - 1] = ‘T‘;bfs(board, row * board[0].length + board[0].length - 1);}}for (int col = 1; col < board[0].length - 1; col++) {if (board[0][col] == ‘O‘) {board[0][col] = ‘T‘;bfs(board, col);}if (board[board.length - 1][col] == ‘O‘) {board[board.length - 1][col] = ‘T‘;bfs(board, (board.length - 1) * board[0].length + col);}}for (int row = 0; row < board.length; row++)for (int col = 0; col < board[0].length; col++) {if (board[row][col] == ‘T‘)board[row][col] = ‘O‘;else if (board[row][col] == ‘O‘)board[row][col] = ‘X‘;}}static public void bfs(char[][] board, int num) {Queue<Integer> queue = new LinkedList<Integer>();queue.add(num);while (!queue.isEmpty()) {num=queue.poll();int row = num / board[0].length;int col = num - row * board[0].length;if (row - 1 >= 0 && board[row - 1][col] == ‘O‘) {board[row - 1][col] = ‘T‘;queue.add(num - board[0].length);}if (row + 1 <= board.length - 1 && board[row + 1][col] == ‘O‘) {board[row + 1][col] = ‘T‘;queue.add(num + board[0].length);}if (col - 1 >= 0 && board[row][col - 1] == ‘O‘) {board[row][col - 1] = ‘T‘;queue.add(num - 1);}if (col + 1 <= board[0].length - 1 && board[row][col + 1] == ‘O‘) {board[row][col + 1] = ‘T‘;queue.add(num + 1);}}}

 

Java for LeetCode 130 Surrounded Regions