牛牛构造等差数列

class Solution:
    def solve(self , n , b ):
        if n<=2:
            return 0
        d_set=set()
        for i in range(-1,2,1):
            for j in range(-1,2,1):
                a0= b[0] + i
                a1= b[1] + j
                d=a1-a0
                d_set.add((a0,d))#记录首项a0和公差d
        res=[]
        for a0_d in d_set:
            r=0
            a0,d=a0_d[0],a0_d[1]
            for k in range(n):
                ak=a0+k*d
                if abs(b[k] - ak)<=1:
                    r+=abs(b[k] - ak)
                else:
                    r=-1
                    break
            res.append(r)
        res_=list(filter(lambda x:x!=-1,res))
        return min(res_) if res_ else -1