130被围绕的区域

题目：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。

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

法一：自己的代码

思路：先绕外围走一圈，将所有与外围相连的岛屿都标记为True，然后把bool数组中位置为False的置为‘X’，为了节省空间可以直接将外围的‘O’,改为‘A’，然后再替换。

from typing import List
class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        r_length = len(board)
        if r_length <= 2:
            return
        c_length = len(board[0])
        if c_length <= 2:
            return
        directions = [[0,1],[0,-1],[1,0],[-1,0]]
        marked = [[False] * c_length for i in range(r_length)]
        # 定义用于标记的函数
        def judge(r,c):
            # 如果没有被遍历过，且为‘O‘则进行遍历
            if not marked[r][c] and board[r][c] == ‘O‘:
                # 先将该位置置为True
                marked[r][c] = True
                # 将与该位置相连的所有‘O’都置为True，表示已经遍历过，
                queue = []
                queue.append((r,c))
                while queue:
                    r,c = queue.pop(0)
                    for p,q in directions:
                        # 注意这里的新位置要用新的变量存储，不可用原先的r,c
                        r_new = r + p
                        c_new = c + q
                        if 0 <= r_new < r_length and 0 <= c_new < c_length and board[r_new][c_new] == ‘O‘ and not marked[r_new][c_new]:
                            # 放入队列，等待遍历
                            queue.append((r_new,c_new))
                            # 遍历过的位置,且为‘O‘的都置为True，
                            marked[r_new][c_new] = True
        # 先遍历最外围一圈,把与最外层相连的岛屿标记为True
        for r in range(r_length):
            if r in [0,r_length-1]:
                for c in range(c_length):
                    judge(r, c)
            else:
                for c in [0,c_length-1]:
                    judge(r, c)
        print(marked)
        # 最后将中间的‘O‘变为’X‘
        for r in range(1, r_length):
            for c in range(1, c_length):
                if not marked[r][c]:
                    board[r][c] = ‘X‘
        print(board)
if __name__ == ‘__main__‘:
    solution = Solution()
    # result = solution.solve([["X","X","X","X"],
    #                          ["X","O","O","X"],
    #                          ["X","X","O","X"],
    #                          ["X","O","X","X"]])
    result = solution.solve([["X","X","X","X"],
                             ["X","O","O","X"],
                             ["X","X","O","X"],
                             ["X","O","X","X"]]

ttt

原文地址：https://www.cnblogs.com/xxswkl/p/12292916.html