题目：

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。

来源：力扣（LeetCode）
链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

示例：

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"]]

解法：

首先分析题目，最后保留的O只可能从边界蔓延得到，所以遍历矩阵四周。

接着把遇到的每个O的坐标入队。

然后BFS，创建和矩阵相同大小的矩阵matrix，用来记录最后的O。取出队头坐标，记录，判断上下左右是否为O且尚未记录，如果是就入队，直到队空。

最后根据matrix修改矩阵。

代码：

class Solution:def solve(self, board: List[List[str]]) -> None:"""Do not return anything, modify board in-place instead."""row = len(board) - 1col = len(board[0]) - 1rh = 1ch = 0flag = 1r = c = 0matrix = [[0] * (col + 1) for _ in range(row + 1)]q = []for _ in range(max(1, 2 * (row + 1 + col + 1) - 4)):if flag == 1:if board[r][c] == 'O' and matrix[r][c] == 0:q.append((r, c))if c == col:r += 1flag = 2else:c += 1elif flag == 2:if board[r][c] == 'O' and matrix[r][c] == 0:q.append((r, c))if r == row:c -= 1flag = 3else:r += 1elif flag == 3:if board[r][c] == 'O' and matrix[r][c] == 0:q.append((r, c))if c == ch:r -= 1flag = 4else:c -= 1else:if board[r][c] == 'O' and matrix[r][c] == 0:q.append((r, c))if r != rh:r -= 1while q:cur = q.pop(0)matrix[cur[0]][cur[1]] = 1try:if board[cur[0] - 1][cur[1]] == 'O' and matrix[cur[0] - 1][cur[1]] == 0:q.append((cur[0] - 1, cur[1]))except IndexError:passtry:if board[cur[0]][cur[1] + 1] == 'O' and matrix[cur[0]][cur[1] + 1] == 0:q.append((cur[0], cur[1] + 1))except IndexError:passtry:if board[cur[0] + 1][cur[1]] == 'O' and matrix[cur[0] + 1][cur[1]] == 0:q.append((cur[0] + 1, cur[1]))except IndexError:passtry:if board[cur[0]][cur[1] - 1] == 'O' and matrix[cur[0]][cur[1] - 1] == 0:q.append((cur[0], cur[1] - 1))except IndexError:passfor r in range(row + 1):for c in range(col + 1):if board[r][c] == 'O' and matrix[r][c] == 0:board[r][c] = 'X'