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

[解题思路]

 board的四个边上的O是无法被X包围的，故如果在四个边上发现O，则这些O应当被保留 从这些O出发继续寻找相邻的O，这些O也是要保留的

 等这些O都标记结束，则剩余的O就应该被改成X

1.使用DFS解，在过大数据集时会挂掉，200*200的矩阵会栈溢出，

public class Solution {
    public static void solve(char[][] board) {
        if (board == null || board.length == 0) {
            return;
        }
        int row = board.length;
        int col = board[0].length;

        // up row
        for (int j = 0; j < col; j++) {
            if (board[0][j] == 'O') {
                dfs(board, 0, j);
            }
        }

        // bottom row
        for (int j = 0; j < col; j++) {
            if (board[row - 1][j] == 'O') {
                dfs(board, row - 1, j);
            }
        }

        // left column
        for (int i = 0; i < row; i++) {
            if (board[i][0] == 'O') {
                dfs(board, i, 0);
            }
        }

        // right column
        for (int i = 0; i < row; i++) {
            if (board[i][col - 1] == 'O') {
                dfs(board, i, col - 1);
            }
        }

        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
                if (board[i][j] == 'P') {
                    board[i][j] = 'O';
                }
            }
        }
    }

    private static void dfs(char[][] board, int i, int j) {
        if(i < 0 || i >= board.length || j < 0 || j > board[0].length || board[i][j] != 'O'){
            return;
        }
        board[i][j] = 'P';

        // up
        if (i - 1 >= 0 && board[i - 1][j] == 'O') {
            dfs(board, i - 1, j);
        }
        // bottom
        if (i + 1 < board.length && board[i + 1][j] == 'O') {
            dfs(board, i + 1, j);
        }
        // left
        if (j - 1 >= 0 && board[i][j - 1] == 'O') {
            dfs(board, i, j - 1);
        }
        // right
        if (j + 1 < board[0].length && board[i][j + 1] == 'O') {
            dfs(board, i, j + 1);
        }

    }
}

 2.BFS

使用BFS来对board进行遍历，如果当前格是'O'，则将该格位置放入到queue中

遍历queue，queue中当前格是'O'时，将其内容改为'P'，则将该格的上下左右都放入到queue中，

public class Solution {
    private Queue<Integer> queue = new LinkedList<Integer>();
    public void solve(char[][] board) {
        if (board.length == 0) {
            return;
        }
        if (board[0].length == 0) {
            return;
        }
        int row = board.length;
        int col = board[0].length;

        // up row
        for (int j = 0; j < col; j++) {
            if (board[0][j] == 'O') {
                bfs(board, 0, j);
            }
        }

        // bottom row
        for (int j = 0; j < col; j++) {
            if (board[row - 1][j] == 'O') {
                bfs(board, row - 1, j);
            }
        }

        // left column
        for (int i = 0; i < row; i++) {
            if (board[i][0] == 'O') {
                bfs(board, i, 0);
            }
        }

        // right column
        for (int i = 0; i < row; i++) {
            if (board[i][col - 1] == 'O') {
                bfs(board, i, col - 1);
            }
        }

        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
                if (board[i][j] == 'P') {
                    board[i][j] = 'O';
                }
            }
        }

    }

    private void fill(char[][] board, int i, int j) {
        int row = board.length;
        int col = board[0].length;
        if (i < 0 || i >= row || j < 0 || j >= col || board[i][j] != 'O')
            return;

        queue.offer(i * col + j);
        board[i][j] = 'P';
    }

    private void bfs(char[][] board, int i, int j) {
        int col = board[0].length;

        fill(board, i, j);
        
        while (!queue.isEmpty()) {
            int cur = queue.poll();
            int x = cur / col;
            int y = cur % col;

            fill(board, x - 1, y);
            fill(board, x + 1, y);
            fill(board, x, y - 1);
            fill(board, x, y + 1);
        }
    }
}

ref:

http://blog.sina.com.cn/s/blog_b9285de20101j1dt.html