极客算法

LeetCode-130.被围绕的区域(Surrounded Regions)


130. 被围绕的区域

给你一个 m x n 的矩阵 board ,由若干字符 ‘X’ 和 ‘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 1:

图片说明

输入:board = [["X","X","X","X"],
              ["X","O","O","X"],
              ["X","X","O","X"],
              ["X","O","X","X"]]
输出:[["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。

示例 2:

输入:board = [[“X”]] 输出:[[“X”]]  

提示:

m == board.length n == board[i].length 1 <= m, n <= 200 board[i][j] 为 ‘X’ 或 ‘O’

来源:力扣(LeetCode)

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

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

宽度优先搜索

O(N)

从任意一个’0’出发,BFS搜索,如果被包围,就设置翻转

from collections import deque
class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        visited = set()
        for i in range(len(board)):
            for j in range(len(board[i])):
                if board[i][j] == 'X' or (i, j) in visited:
                    continue

                self.helper(i, j, board, visited)

    def helper(self, i : int, j : int, board: List[List[str]], visited: set):
        
        directions = [-1, 0, 1, 0, -1] # x, y合一方向数组
        start = (i, j)
        queue = deque([start])
        visited.add(start)
        captured = True
        booty = [start]
        
        
        while len(queue) > 0:
            x, y = queue.popleft()
            
            for k in range(4):
                neighbor = (x + directions[k], y + directions[k + 1])
                if 0 <= neighbor[0] < len(board) and 0 <= neighbor[1] < len(board[i]):
                    if neighbor not in visited:
                        visited.add(neighbor)
                        if board[neighbor[0]][neighbor[1]] == 'O':
                            queue.append(neighbor)
                            booty.append(neighbor)
                else:
                    captured = False

        if captured:
            for x, y in booty:
                board[x][y] = 'X'

从四个边开始BFS,记录不能翻转的,然后统一处理

from collections import deque
class Solution:
    def solve(self, board: List[List[str]]) -> None:       
        """
        Do not return anything, modify board in-place instead.
        """
        visited = set()

        for i in range(len(board)):
            self.bfs(board, visited, i, 0) # 第一列
            self.bfs(board, visited, i, len(board[i]) - 1) # 最后一列

        for j in range(1, len(board[0]) - 1):
            self.bfs(board, visited, 0, j) # 第一行
            self.bfs(board, visited, len(board) - 1, j) # 最后一行

        for i in range(len(board)):
            for j in range(len(board[i])):
                if board[i][j] == 'O' and (i, j) not in visited:
                    board[i][j] = 'X'

    def bfs(self, board, visited, i, j):

        if board[i][j] == 'X' or (i, j) in visited:
            return

        directions = [-1, 0, 1, 0, -1]
        queue = deque([(i, j)])
        visited.add((i, j))

        while(len(queue) > 0):
            x, y = queue.popleft()
            
            for k in range(4):
                neighbor = (x + directions[k], y + directions[k + 1])

                if 0 <= neighbor[0] < len(board) and 0 <= neighbor[1] < len(board[i]) and neighbor not in visited and board[neighbor[0]][neighbor[1]] == 'O':
                    visited.add(neighbor)
                    queue.append(neighbor)

深度优先搜索

也可以用深度优先搜索,只是广度优先比较简单

并查集

定义一个超级节点代表’自由’,把所有从边遍历的节点’0’,都与这个自由节点相连, 最后没有与’自由’节点相连的同一翻转成’X’

–End–


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