SurroundedRegions

被围绕的区域

• 题目介绍
• 题目解法

题目介绍

被围绕的区域

给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

示例 1:

输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。 

示例 2:

输入：board = [["X"]]
输出：[["X"]]

提示：

• m == board.length
• n == board[i].length
• 1 <= m, n <= 200
• board[i][j] 为 'X' 或 'O'

题目解法

 int m;
    int n;
    public void solve(char[][] board) {

        m = board.length; // 行
        n = board[0].length; // 列
if (m == 1 || n == 1) {
            return;
        }
        for (int i = 0; i < n; i++) {
            dfs(board, 0, i);
            dfs(board, m - 1, i);
        }
        for (int i = 1; i < m - 1; i++) {
            dfs(board, i, 0);
            dfs(board, i, n - 1);
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'A') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }

    }

    private void dfs(char[][] board, int x, int y) {
        if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || board[x][y] != 'O') {
            return;
        }
        board[x][y] = 'A';
        dfs(board, x + 1, y);
        dfs(board, x - 1, y);
        dfs(board, x, y + 1);
        dfs(board, x, y - 1);
    }

打印：

思路：

思路上， 注意dfs的结束条件是不等于O，因为边界可能被来回dfs，O会被标记为A，不会来回标记。