最少的python代码

思路：
遍历每个位置的情况，当遇到?时，看它的周围有没有地雷，有的话count+=1，然后把count值放到另一个矩阵res中。

n,m = map(int,input().split())
s = [list(input()) for _ in range(n)]
res = [['*'] *m for _ in range(n)]
direction = [(0,1),(0,-1),(1,0),(-1,0),(1,1),(1,-1),(-1,1),(-1,-1)]
count = 0
for i in range(n):
    for j in range(m):
        if s[i][j]=='?':
            count = 0
            for dx,dy in direction:
                x = i + dx
                y = j + dy
                if 0<=x<n and 0<=y<m and s[x][y]=="*":
                    count+=1
            res[i][j] = str(count)

for i in res:
    print("".join(i))