面试150 生命游戏

思路

深度优先遍历的思路，对当前细胞八个方向进行遍历，统计存活的细胞，并通过标记修改board数组，如果细胞死亡则标记为-1，细胞复活标记为2。再次遍历数组，然后根据标记修改数组即可

class Solution:
    def gameOfLife(self, board: List[List[int]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        m=len(board)
        n=len(board[0])
        directions=[(-1,0),(0,-1),(0,1),(1,0),(1,1),(-1,1),(-1,-1),(1,-1)]

        def dfs(i,j):
            count=0
            for x,y in directions:
                new_x=x+i 
                new_y=y+j 
                if 0<=new_x<m and 0<=new_y<n and abs(board[new_x][new_y])==1:
                    count+=1
            return count

        for i in range(m):
            for j in range(n):
                live=dfs(i,j)
                if board[i][j]==1 and live<2:
                    board[i][j]=-1
                elif board[i][j]==1 and live>3:
                    board[i][j]=-1
                elif board[i][j]==0 and live==3:
                    board[i][j]=2
        for i in range(m):
            for j in range(n):
                if board[i][j]<0:
                    board[i][j]=0
                elif board[i][j]>0:
                    board[i][j]=1                

值得注意的是，abs(board[new_x][new_y])==1，是为了正确统计当前细胞的邻居原本是"活"的数量。因为在原地修改的时候-1表示原来是活的，下一个状态要变成死