LeetCode | 0130. Surrounded Regions被围绕的区域【Python】

LeetCode 0130. Surrounded Regions被围绕的区域【Medium】【Python】【DFS】

Problem

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.

问题

力扣

给定一个二维的矩阵，包含 'X' 和 'O'（字母 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:

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

运行你的函数后，矩阵变为：

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。

思路

DFS

从反面来想。
先找到边界（即上下左右四条边）上的 "O"，先变为 "*"。
再遍历 board 矩阵，将 "O" 变为 "X"，再将 "*" 变为 "O"。
这样问题就变成从边界找连通的 "O"，就可以用 DFS 了。

时间复杂度: O(m*n)，m 是行数，n 是列数。
空间复杂度: O(1)，因为是在 board 矩阵基础上进行改动。

Python3代码

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        if not board:
            return None
        
        m, n = len(board), len(board[0])
        directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]

        def dfs(i, j):
            if 0 <= i < m and 0 <= j < n and board[i][j] == "O":  # change "O" to "*"
                board[i][j] = "*"
                for k in range(4):
                    dfs(i + directions[k][0], j + directions[k][1])
        
        # search left and right
        for i in range(m):
            dfs(i, 0)
            dfs(i, n - 1)
        
        # search up and down
        for i in range(n):
            dfs(0, i)
            dfs(m - 1, i)
        
        for i in range(m):
            for j in range(n):
                if board[i][j] == "O":  # change "O" to "X"
                    board[i][j] = "X"
                elif board[i][j] == "*":  # change "*" to "O"
                    board[i][j] = "O"

代码地址

GitHub链接