有没有什么方法可以减少这个cod的时间和代码

Question

我写了一个运行时间为O（n^2）的程序，有没有可能在不使用嵌套的forloops和内置函数的情况下降低代码和复杂度，这是一个匹配的整数问题，输出是[4,5]

a = 12345
b = 49085
l1 = []
l2 = []
l_match = []
while(a != 0):
    c = a % 10
    l1.append(c)
    a //= 10
while(b!= 0):
    d = b % 10
    l2.append(d)
    b //= 10
for i in range(len(l1)):
    for j in range(len(l2)):
        if l1[i] == l2[j]:
            match = l1[i]
            l_match.append(match)
print(l_match)

Answer 1

时间复杂度

def find_chars_order_n(value1,value2):
  hsh = {}
  resp = set()     #Set is used to have unique Characters
  if str(value1) and str(value2):
    str1 = list(str(value1))
    str2 = list(str(value2))
  for elm in str2:
    hsh[elm] = elm
  for elm in str1:
    if hsh[elm]:
      resp.add(elm)
  # resp = str1 & str2 #using set intersection
  if resp:
    return list(resp))
  else:
    return None

if __name__ == "__main__":
  print(find_chars_order_n(1400,4011))