130-被围绕的区域

LeetCode 130. 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。

method

先把边缘的联通O区域都变成A，后面再遍历的时候，矩阵中的O就都是被围绕的区域，直接改成X，同时也把A还原回O

int m, n;
void dfs(vector<vector<char>>& board, int x, int y) {
    if (x < 0 || x >= m || y < 0 || y >= n || 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);
}
void solve(vector<vector<char>>& board) {
    m = board.size(), n = board[0].size();
    for (int i = 0; i < m; i++) {
        dfs(board, i, 0);
        dfs(board, i, n - 1);
    }
    for (int j = 0; j < n; j++) {
        dfs(board, 0, j);
        dfs(board, m - 1, j);
    }
    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';
        }
    }
}