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  • 130被围绕的区域

    from typing import List
    # 这道题看了大佬写的代码,经过自己的理解写出来了。
    # 从最外围的四周找有没有为O的,如果有的话就进入深搜函数,然后深搜遍历
    # 判断上下左右的位置是否为O
    class Solution:
    def solve(self, board: List[List[str]]) -> None:
    # 判断是否为空列表
    if not board or not board[0] : return
    # 求出列表的有几行和有几列
    row = len(board)
    col = len(board[0])
    # 深搜,当边界为O进行判断
    def dfs(index1,index2):
    # 首先将O变成B,之后再变回来。
    board[index1][index2] = 'B'
    # 遍历上下左右
    for x,y in [(0,1),(0,-1),(1,0),(-1,0)]:
    print(x,y)
    # 注意这里不能改变index1和index2的值,因为后边还要用。
    i = index1 + x
    j = index2 + y
    # 如果当前位置的上下左右不出边界而且位置上为O就,进行深搜
    # 注意这里边界上边不能判断,
    if 1 <= i < row and 1 <= j < col and board[i][j] == "O":
    print("111")
    dfs(i,j)
    # 判断第一行和最后一行上的边界值
    for index1 in range(col):
    if board[0][index1] == "O":
    dfs(0,index1)
    if board[row - 1][index1] == 'O':
    dfs(row - 1,index1)
    # 判断第一列和最后一列的边界值
    for index2 in range(row):
    if board[index2][0] == "O":
    dfs(index2,0)
    if board[index2][col - 1] == 'O':
    dfs(index2,col - 1)
    # 将O全部改写成X,将之前写的B也改写成O
    for index1 in range(row):
    for index2 in range(col):
    if board[index1][index2] == "O":
    board[index1][index2] = "X"
    if board[index1][index2] == "B":
    board[index1][index2] = 'O'

    A = Solution()
    print(A.solve([["O","X","X","O","X"],["X","O","O","X","O"],["X","O","X","O","X"],["O","X","O","O","O"],["X","X","O","X","O"]])
    )
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  • 原文地址:https://www.cnblogs.com/cong12586/p/13399237.html
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