題目要求:
https://leetcode-cn.com/problems/surrounded-regions/
class Solution:
def solve(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
# 如果是空數組,直接返回
# Leetcode總搞這種邊邊角角的輸入
if not board: return
# 計算數組長寬
row = len(board)
col = len(board[0])
# 如果長度或者寬度中一個小於3的話也不用算了,直接返回
if row < 3 or col < 3: return
# DFS函數
def dfs(i, j):
# 如果i,j中有一個越界或者遇到了X則不繼續掃描
if i < 0 or j < 0 or i >= row or j >= col or board[i][j] != 'O':
return
# 否則把數組中的O變成#,意思是這個O和邊緣是連通的
board[i][j] = '#'
# 之後從當前座標開始上下左右進行遞歸搜索
dfs(i - 1, j)
dfs(i + 1, j)
dfs(i, j - 1)
dfs(i, j + 1)
for i in range(row):
# 搜索第一行和最後一行
dfs(i, 0)
dfs(i, col - 1)
for i in range(col):
# 搜索第一列和最後一列
dfs(0, i)
dfs(row - 1, i)
# 全部搜索完畢後:
# X X X X
# X X O X
# X O X X
# X O X X
# 變爲:
# X X X X
# X X O X
# X # X X
# X # X X
# 之後再將所有的#變成O,O變成X就可以了
for i in range(0, row):
for j in range(0, col):
if board[i][j] == 'O':
board[i][j] = 'X'
if board[i][j] == '#':
board[i][j] = 'O'