Leetcode - 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
[分析] 使用DFS在大数据时会stackoverflow, 因此选择BFS。边界上的O不会变为 X，因此对边界上各元素进行BFS，BFS搜索时遇到的 O 均从边界 O可达，因此将不翻转，将已搜索过的 O 标记为特殊字符比如 M 以避免重复搜索死循环。搜索完成后所有剩余的 O 均需翻转，并将所有 M 恢复为 O。 BFS搜索时队列中维护的是所有未展开BFS的 O，队列记录这些 O的位置。

public class Solution {
    public void solve(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) return;
        int rows = board.length, cols = board[0].length;
        for (int i = 0; i < rows; i++) {
            bfs(i, 0, board);
            bfs(i, cols - 1, board);
        }
        for (int j = 0; j < cols; j++) {
            bfs(0, j, board);
            bfs(rows - 1, j, board);
        }
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (board[i][j] == 'M') 
                    board[i][j] = 'O';
                else if (board[i][j] == 'O') 
                    board[i][j] = 'X';
            }
        }
    }
    private void bfs(int i, int j, char[][] board) {
        LinkedList<Integer> queue = new LinkedList<Integer>();
        visit(i, j, board, queue);
        while (!queue.isEmpty()) {
            int idx = queue.poll();
            int r = idx / board[0].length;
            int c = idx % board[0].length;
            visit(r + 1, c, board, queue);
            visit(r - 1, c, board, queue);
            visit(r, c - 1, board, queue);
            visit(r, c + 1, board, queue);
        }
    }
    private void visit(int i, int j, char[][] board, LinkedList<Integer> queue) {
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != 'O')
            return;
        board[i][j] = 'M';
        queue.offer(i * board[0].length + j);
    }
}