Surrounded Regions

最新推荐文章于 2022-06-01 17:41:20 发布

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最新推荐文章于 2022-06-01 17:41:20 发布

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分类专栏： python Leetcode dfs union-find

本文链接：https://blog.youkuaiyun.com/github_15549139/article/details/60961119

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本文探讨了如何通过深度优先搜索(DFS)和并查集(Union-find)解决二维棋盘中被X包围的O型区域捕获问题，并提供两种方法的具体实现代码。

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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.

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

用了DFS，发现会内存溢出。思路如下，对于边际上的每一个’O’，找到他们的连通分量，设置为’S’。之后将’O’设置为’X’，将’S’设置为’O’。代码如下：

class Solution(object):
    def dfs(self, i, j, board):
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]):
            return

        if board[i][j] == 'O':
            board[i][j] = 'q'
            self.dfs(i, j+1, board)
            self.dfs(i, j-1, board)
            self.dfs(i-1, j, board)
            self.dfs(i+1, j, board)

    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        if len(board) != 0:
            for i in range(len(board)):
                if i == 0 or i == len(board) - 1:
                    for j in range(len(board[0])):
                        if board[i][j] == 'O':
                            self.dfs(i, j, board)
                else:
                    for j in [0, len(board[0]) - 1]:
                        if board[i][j] == 'O':
                            self.dfs(i, j, board)

            for i in range(len(board)):
                for j in range(len(board[0])):
                    if board[i][j] == 'O':
                        board[i][j] = 'X'

            for i in range(len(board)):
                for j in range(len(board[0])):
                    if board[i][j] == 'q':
                        board[i][j] = 'O'

这样想来，这一类题目还是用Union-find让人觉得更安心一点。

class UF(object):
    def union(self, a, b, tmp_board, size):
        p = self.root(a, tmp_board)
        q = self.root(b, tmp_board)
        if p == q:
            return
        if size[p] > size[q]:
            tmp_board[q] = p
            size[p] += size[q]
        else:
            tmp_board[p] = q
            size[q] += size[p]

    def root(self, a, tmp_board):
        while(tmp_board[a] != a):
            tmp_board[a] = tmp_board[tmp_board[a]]
            a = tmp_board[a]
        return a

    def connected(self, a, b, tmp_board):
        return self.root(a, tmp_board) == self.root(b, tmp_board)

class Solution(object):
    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        m = len(board)
        uf = UF()
        if m == 0:
            return
        n = len(board[0])
        tmp_board = [i for i in range(m*n+1)]
        size = [1] * (m*n+1)

        for i in range(m):
            for j in range(n):
                if board[i][j] == 'O':
                    if j+1 < n:
                        if board[i][j+1] == 'O':
                            uf.union(i*n+j, i*n+j+1, tmp_board, size)
                    if i+1 < m:
                        if board[i+1][j] == 'O':
                            uf.union((i+1)*n+j, i*n+j, tmp_board, size)

        for i in range(m):
            if i == 0 or i == m-1:
                for j in range(n):
                    uf.union(m*n, i*n+j, tmp_board, size)
            else:
                for j in [0, n-1]:
                    uf.union(m*n, i*n+j, tmp_board, size)

        for i in range(m):
            for j in range(n):
                if not uf.connected(i*n+j, m*n, tmp_board):
                    board[i][j] = 'X'