leetcode_130 被围绕的区域

要求

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'

核心代码

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        for i in range(len(board)):
            for j in range(len(board[i])):
                if i in [0,len(board) - 1] or j in [0,len(board[i]) - 1]:
                    if board[i][j] == "O":
                        self.saved(board,i,j)
        for i in range(len(board)):
            for j in range(len(board[i])):
                if board[i][j] == "O":
                    board[i][j] = "X"
        for i in range(len(board)):
            for j in range(len(board[i])):
                if board[i][j] == "A":
                    board[i][j] = "O"
    
    def saved(self,matrix,i,j):
        matrix[i][j] = "A"
        for di in [[0,1],[0,-1],[1,0],[-1,0]]:
            x,y = i + di[0],j + di[1]
            if 0 <= x < len(matrix) and 0 <= y < len(matrix[i]):
                if matrix[x][y] == "O":
                    self.saved(matrix,x,y)

解题思路：第一步：找到边缘的O，将它和与它临接的O都变为A；第二步：没有变为A的O都是应该变为X的；第三步：将A变回原来的O。第一个值得注意的部分：我们可以看到第一个部分i in [0,len(board) - 1]就是边缘的部分；第二个值得注意的部分：我们关注saved函数，[[0,1],[0,-1],[1,0],[-1,0]]恰好是选中点的上下左右的点。