给定一个二维的矩阵，包含 'X' 和 'O'（字母 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:X X X XX O O XX X O XX O X X运行你的函数后，矩阵变为：X X X XX X X XX X X XX 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。

/* * @lc app=leetcode id=130 lang=javascript * * [130] Surrounded Regions */// 将O以及周边的O转化为Afunction mark(board, i, j, rows, cols) {  if (i < 0 || i > rows - 1 || j < 0 || j > cols - 1 || board[i][j] !== "O")    return;  board[i][j] = "A";  mark(board, i + 1, j, rows, cols);  mark(board, i - 1, j, rows, cols);  mark(board, i, j + 1, rows, cols);  mark(board, i, j - 1, rows, cols);}/** * @param {character[][]} board * @return {void} Do not return anything, modify board in-place instead. */var solve = function (board) {  const rows = board.length;  if (rows === 0) return [];  const cols = board[0].length;  for (let i = 0; i < rows; i++) {    for (let j = 0; j < cols; j++) {      if (i === 0 || i == rows - 1 || j === 0 || j === cols - 1) {        mark(board, i, j, rows, cols);      }    }  }  for (let i = 0; i < rows; i++) {    for (let j = 0; j < cols; j++) {      if (board[i][j] === "O") {        board[i][j] = "X";      } else if (board[i][j] === "A") {        board[i][j] = "O";      }    }  }  return board;};

class Solution:    def solve(self, board: List[List[str]]) -> None:        """        Do not return anything, modify board in-place instead.        """        # 如果数组长或宽小于等于2，则不需要替换        if len(board) <= 2 or len(board[0]) <= 2:            return        row, col = len(board), len(board[0])        def dfs(i, j):            """            深度优先算法，如果符合条件，替换为A并进一步测试，否则停止            """            if i < 0 or j < 0 or i >= row or j >= col or board[i][j] != 'O':                return            board[i][j] = 'A'            dfs(i - 1, j)            dfs(i + 1, j)            dfs(i, j - 1)            dfs(i, j + 1)        # 从外围开始        for i in range(row):            dfs(i, 0)            dfs(i, col-1)        for j in range(col):            dfs(0, j)            dfs(row-1, j)        # 最后完成替换        for i in range(row):            for j in range(col):                if board[i][j] == 'O':                    board[i][j] = 'X'                elif board[i][j] == 'A':                    board[i][j] = 'O'

class Solution {    int M, N, dirs[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};    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] = '#';        for (auto &dir : dirs) dfs(board, x + dir[0], y + dir[1]);    }public:    void solve(vector<vector<char>>& board) {        if (board.empty() || board[0].empty()) return;        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 (auto &row : board) {            for (auto &cell : row) {                cell = cell == '#' ? 'O' : 'X';            }        }    }};