LeetCode-130. Surrounded Regions

最新推荐文章于 2022-02-28 07:33:26 发布

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最新推荐文章于 2022-02-28 07:33:26 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_17753903/article/details/82937456

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本文介绍了一个算法问题，即在一个包含'O'和'X'的二维棋盘中，如何找出并翻转所有被'X'包围的'O'区域。通过深度优先搜索查找所有相连的'O'，并判断其是否位于边界。如果'O'区域不接触边界，则将其全部翻转为'X'。

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0.原题

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example:

X X X X
X O O X
X X O X
X O X X

After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

Explanation:

Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

1.代码

class Solution:
    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        self.rows = len(board)
        if self.rows:
            self.columns = len(board[0])
            self.visited = [[False for _ in range(self.columns)]for _ in range(self.rows)]
            for r in range(self.rows):
                for c in range(self.columns):
                    if board[r][c] == "O" and not self.visited[r][c]:
                        coordinate_dict = {'rows':[],'columns':[]}
                        self.visit(board,r,c,coordinate_dict)
                        self.flip(board,coordinate_dict)
                        
    def visit(self,board,row,column,coordinate_dict):
        if board[row][column] == 'X':
            return 0
        else:
            if not self.visited[row][column]:
                coordinate_dict['rows'].append(row)
                coordinate_dict['columns'].append(column)
                self.visited[row][column] = True
                if row - 1 >= 0 :
                    self.visit(board,row-1,column,coordinate_dict)
                if row + 1 < self.rows:
                    self.visit(board,row+1,column,coordinate_dict)
                if column - 1 >= 0:
                    self.visit(board,row,column-1,coordinate_dict)
                if column + 1 < self.columns:
                    self.visit(board,row,column+1,coordinate_dict)

    def flip(self, board, coordinate_dict):
        min_row = min(coordinate_dict['rows'])
        min_column = min(coordinate_dict['columns'])
        max_row = max(coordinate_dict['rows'])
        max_column = max(coordinate_dict['columns'])
        if min_row == 0 or min_column == 0 or max_row == self.rows-1 or max_column == self.columns-1:
            return 0
        else:
            for (row, column) in zip(coordinate_dict['rows'], coordinate_dict['columns']):
                board[row][column] = "X"


if __name__ == "__main__":
    board = [["X","O","X","O","X","O"],["O","X","O","X","O","X"],["X","O","X","O","X","O"],["O","X","O","X","O","X"]]
    board2 = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
    solution = Solution()
    solution.solve(board2)
    print(board)

2.思路

首先查找所有相连的'O'，查找的方法与“岛屿”问题类似，不在赘述。

https://blog.youkuaiyun.com/qq_17753903/article/details/82933759

然后，判断连接在一起的‘O’，是否有在边界的：

如果有，则不用翻转；

如果没有，证明被'X'所包围，将所有的'O'翻转为'X'。