[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

Solution:

通过对四条边上的点进行遍历来判断边缘上是否有‘O’，

1. 如果四条边上都没有‘O’，说明'O'(如果有)都被'X'给包围了，遍历整个二维数组直接将‘O’改为‘X’即可。

2. 如果发现了‘O’，沿着该‘O’的上下左右方向，分别判断是否是‘O’，如果是，加入队列。

public class Solution {
    public void solve(char[][] board) {
        if (board == null || board.length == 0)
            return;
        if (board[0].length == 0)
            return;
        int row = board.length;
        int col = board[0].length;
        for (int i = 0; i < row; ++i) {
            bfs(board, i, 0);
            bfs(board, i, col - 1);
        }
        for (int i = 1; i < col - 1; ++i) {
            bfs(board, 0, i);
            bfs(board, row - 1, i);
        }
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (board[i][j] == 'O')
                    board[i][j] = 'X';
            }
        }
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (board[i][j] == 'D')
                    board[i][j] = 'O';
            }
        }
    }

    private void bfs(char[][] board, int x, int y) {
        if(board[x][y]!='O')
            return;
        board[x][y]='D';
        Queue<Integer> queue=new LinkedList<Integer>();
        queue.add(x*board[0].length+y);
        while (!queue.isEmpty()) {
            int code=queue.poll();
            int row=code/board[0].length;
            int col=code%board[0].length;
            if(row-1>=0&&board[row-1][col]=='O'){
                queue.add((row-1)*board[0].length+col);
                board[row-1][col]='D';
            }
            if(row+1<board.length&&board[row+1][col]=='O'){
                queue.add((row+1)*board[0].length+col);
                board[row+1][col]='D';
            }
            if(col-1>=0&&board[row][col-1]=='O'){
                queue.add(row*board[0].length+col-1);
                board[row][col-1]='D';
            }
            if(col+1<board[0].length&&board[row][col+1]=='O'){
                queue.add(row*board[0].length+col+1);
                board[row][col+1]='D';
            }
        }
        
    }
}