130. 被围绕的区域

思路

• 从边界上的O开始DFS，把搜到的O都置为1。

• 扫一遍矩阵，把所有的O都置为X。

• 再扫一遍矩阵，把所有的1都置为O。

方法一： DFS

class Solution {
    public void solve(char[][] board) {
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (i == 0 || i == board.length - 1
                    || j == 0 || j == board[0].length - 1) {
                    dfs(board, i, j);
                }
            }
        }
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == 'O') {//把内部封闭的o替换成x
                    board[i][j] = 'X';
                }
                if (board[i][j] == '@') {//联通的恢复
                    board[i][j] = 'O';
                }
            }
        }
    }

    public void dfs(char[][] board, int i, int j) {
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != 'O') {
            return;
        }
        board[i][j] = '@';//与边界连通的‘O’设置为‘@’
        dfs(board, i + 1, j);
        dfs(board, i - 1, j);
        dfs(board, i, j + 1);
        dfs(board, i, j - 1);

    }
}