leetcode @python 130. Surrounded Regions

题目链接

https://leetcode.com/problems/surrounded-regions/

题目原文

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

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

For 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

题目大意

给出一个铺满O或者X的区域，将被X围住的O都换成X

解题思路

使用BFS从边上开始搜索，如果是'O'，那么搜索'O'周围的元素，并将'O'置换为'D'，这样每条边都DFS或者BFS一遍。而内部的'O'是不会改变的。这样下来，没有被围住的'O'全都被置换成了'D'，被围住的'O'还是'O'，没有改变。然后遍历一遍，将'O'置换为'X'，将'D'置换为'O'。

代码

class Solution(object):
    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """

        def fill(x, y):
            if x < 0 or x > m - 1 or y < 0 or y > n - 1 or board[x][y] != 'O':
                return
            queue.append((x, y))
            board[x][y] = 'D'

        def bfs(x, y):
            if board[x][y] == 'O':
                fill(x, y)
            while queue:
                cur = queue.pop(0)
                i = cur[0]
                j = cur[1]
                fill(i + 1, j)
                fill(i - 1, j)
                fill(i, j + 1)
                fill(i, j - 1)

        if len(board) == 0:
            return
        m = len(board)
        n = len(board[0])
        queue = []

        for i in range(n):
            bfs(0, i)
            bfs(m - 1, i)
        for j in range(1, m - 1):
            bfs(j, 0)
            bfs(j, n - 1)

        for i in range(m):
            for j in range(n):
                if board[i][j] == 'D':
                    board[i][j] = 'O'
                elif board[i][j] == 'O':
                    board[i][j] = 'X'