思路:
這題的關鍵在於不能一個一個的更新,每一個位置都要按照原來的周圍環境進行更新,也可以理解爲同時更新。
用0作爲界限來區分活細胞還是死細胞。一次遍歷,特殊標記,再次遍歷復原即可。
特殊的點爲:原來是活的,更新後死了,原來是死的,更新後活了。
class Solution:
def gameOfLife(self, board: List[List[int]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
row = len(board)
col = len(board[0])
for i in range(row) :
for j in range(col):
count = self.dfs(i,j,board)
if board[i][j] > 0: #活細胞,dfs後死了的用2標記,還是活的不管
if count < 2 or count > 3:
board[i][j] = 2
else: #死細胞,dfs後變活的用-1標記,還是死的不管
if count == 3:
board[i][j] = -1
#特殊標記的再變回來
for i in range(row):
for j in range(col):
if board[i][j] == 2:
board[i][j] = 0
if board[i][j] == -1:
board[i][j] = 1
return board
def dfs(self,i,j,board):
dx = [-1,0,1,-1,1,-1,0,1]
dy = [-1,-1,-1,0,0,1,1,1]
count = 0
for k in range(8):
x = i + dx[k]
y = j + dy[k]
if x < 0 or x >= len(board) or y < 0 or y >= len(board[0]):
continue
elif board[x][y] > 0:
count += 1
return count