leetcode130.被围绕的区域

首先对边界上每一个'O'做深度优先搜索，将与其相连的所有'O'改为'P'。然后遍历矩阵，将矩阵中所有'O'改为'X',将矩阵中所有P'变为'O' 。

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        if not board or len(board) == 1 or len(board[0]) == 1:
            return
        m, n = len(board), len(board[0])
        dx = [1, -1, 0, 0]
        dy = [0, 0, 1, -1]
        def dfs(x,y):
            for k in range(4):
                xx = x + dx[k]
                yy = y + dy[k]
            
                if 0 <= xx <m and 0 <= yy < n:
                    if board[xx][yy] == 'O':
                        board[xx][yy] = 'P'   #给走过的点染色
                        dfs(xx, yy)
        # 以下在从四周往里找相邻的所有点
        i = 0
        for j in range(n): # 上边
            if board[i][j] == 'O':
                board[i][j] = 'P'
                dfs(i, j)
        i = m - 1
        for j in range(n): # 下边
            if board[i][j] == 'O':
                board[i][j] = 'P'
                dfs(i, j)
        j = 0
        for i in range(m): # 左边
            if board[i][j] == 'O':
                board[i][j] = 'P'
                dfs(i, j)        
        j = n - 1
        for i in range(m): # 右边
            if board[i][j] == 'O':
                board[i][j] = 'P'
                dfs(i, j)
        # 染色回来
        for i in range(m):
            for j in range(n):
                if board[i][j] == 'P':
                    board[i][j] = 'O'
                elif board[i][j] == 'O':
                    board[i][j] = 'X'
        return