周边地区Python
假设我们有一个包含X和O的2D板。捕获X包围的所有区域。通过将该包围的区域中的所有O更改为X来捕获区域。
运行后输出
为了解决这个问题，我们将遵循以下步骤-
如果不存在板，则返回空白板
对于范围从0到行数的i-1-
如果board[i，0]='O'，则make_one(board,i,0)
如果board[i，行长-1]='O'，则make_one（board，i，行长-1）
对于i在0到cols数量的范围内-1−
如果board[0，i]='O'，则make_one(board,0,i)
如果board[行数-1，i]='O'，则make_one（board，行数-1，i）
对于0到行数范围内的i
如果board[i，j]='O'，则board[i，j]='X'，否则为1，board[i，j]='O'
对于范围从0到列数的j
make_one将像-
如果i<0或j<0或i>=行数或j>=列数或board[i，j]='X'或board[i，j]='1'，则返回
板[i，j]：=1
呼叫make_one（voard，i+1，j），make_one(voard,i-1,j)，make_one（voard，i，j+1），make_one(voard,i,j-1)
示例
让我们看下面的实现以更好地理解-
class Solution(object):
def solve(self, board):
"""
:type board: List[List[str]]
:rtype: None Do not return anything, modify board in-place instead.
"""
if not board:
return board
for i in range(len(board)):
if board[i][0]=='O':
self.make_one(board,i,0)
if board[i][len(board[0])-1] == 'O':
self.make_one(board,i,len(board[0])-1)
for i in range(len(board[0])):
if board[0][i]=='O':
self.make_one(board,0,i)
if board[len(board)-1][i] == 'O':
self.make_one(board,len(board)-1,i)
for i in range(len(board)):
for j in range(len(board[i])):
if board[i][j]=='O':
board[i][j]='X'
elif board[i][j]=='1':
board[i][j]='O'
def make_one(self, board,i,j):
if i<0 or j<0 or i>=len(board) or j>=len(board[0]) or board[i][j]=='X' or board[i][j]=='1':
return
board[i][j]='1'
self.make_one(board,i+1,j)
self.make_one(board,i-1,j)
self.make_one(board,i,j+1)
self.make_one(board,i,j-1)输入项
[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出结果
[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]