題目如下：

解題思路：因為本題要求的是直接在輸入的數組上面修改，而且細胞的生死轉換是一瞬間的事情，所以需要引入兩個中間狀態，待死和待活。這兩個中間狀態用於在四條規則判斷的時候是“活”和“死”，但是呈現在最終的結果是“死”和“活”。

代碼如下：

class Solution(object):
    def getLiveCount(self,board,x,y):
        l = [(0,-1),(0,1),(-1,0),(1,0),(-1,-1),(1,1),(-1,1),(1,-1)]
        count = 0
        for i in l:
            if
 
 board[x+i[0]][y+i[1]] == 1 or board[x+i[0]][y+i[1]] == 4:
                count += 1
        return count
    def gameOfLife(self, board):
        """
        :type board: List[List[int]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        # 2 means boundary
 

        # 3 means from dead to live
        # 4 means from live to dead
        for i in board:
            i.append(2)
            i.insert(0,2)
        l = [2] * len(board[0])
        board.append(l[::])
        board.insert(0,l[::])

        for i in range(1,len(board)-1):
            for j in range(1,len(board[i])-1):
 

                count = self.getLiveCount(board,i,j)
                if board[i][j] == 0 and count == 3:
                    board[i][j] = 3
                elif board[i][j] == 1 and (count < 2 or count > 3):
                    board[i][j] = 4
        for i in range(1,len(board)-1):
            for j in range(1,len(board[i])-1):
                if board[i][j] == 3:
                    board[i][j] = 1
                elif board[i][j] == 4:
                    board[i][j] = 0
        del board[0]
        del board[-1]
        for i in board:
            del i[0]
            del i[-1]

【leetcode】289. Game of Life