题解 | #进制转换#
进制转换
http://www.nowcoder.com/practice/2cc32b88fff94d7e8fd458b8c7b25ec1
总的思路是从后往前处理数值
class Solution:
def solve(self , M: int, N: int) -> str:
K=abs(M)
S=""
n=0
L=['0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F']
F = '-' if M<0 else ''
while K>=N**n:
i=K%(N**(n+1))
j=i//(N**n)
S+=L[j]
K-=i
n+=1
return F+S[::-1]
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