给定任意个正整数，请编写一个函数，求出这些整数的最小公倍数。
【示例】
输入：2,3,4
输出：12
解释：根据数学常识，2，3，4的最小公倍数是12。
题目难度：中等
题目来源：codewars-Least Common Multiple
def solution(*args)-> int:
# your code here
assert solution(2,3,4) == 12
assert solution(2,5) == 10
assert solution(9) == 9
assert solution(2, 4, 27, 8, 3, 9, 24, 216, 288, 51288) == 1846368
def solution(*args) -> int:
# your code here
if 0 in args:
return 0
elif len(args) == 0:
return 1
elif len(args) == 1:
return args[0]
else:
greater = max(args)
while True:
if len(list(filter(lambda x: greater % x == 0, args))) == len(args):
lcm = greater
break
greater += 1
return lcm
assert solution(0, 1, 2) == 0
assert solution(2, 3, 4) == 12
assert solution(2, 5) == 10
assert solution(2, 4) == 4
assert solution(9) == 9
assert solution() == 1
2，4
def solution(*args) -> int:
if len(args) == 1:
return args[0]
sum = max(args)
for a in args:
if sum % a == 0:
continue
sum *= a
min_cm = 0
while (sum > 0):
flag = True
for a in args:
if sum % a !=0:
flag = False
break
if flag:
min_cm = sum
sum -= 1
return min_cm
assert solution(2,3,4) == 12
assert solution(2,5) == 10
assert solution(9) == 9
assert solution(2,4) == 4
嗦嘎 感谢提醒
def solution(*args) -> int:
size = len(args)
idx = 1
i = args[0]
f = 0 # 给定初始化结果
if size == 1: # 如果长度为1，则最大公倍数为i
return i
while idx < size:
j = args[idx]
b = i if i < j else j
a = i if i > j else j
while a != 0: # 使用欧几里得公式求两个数的最大公约数 b
a, b = b % a, a
f = int(i * j / b) # 最小公倍数=两数相乘/最大公约数
i = f # 阶段性得出来最大公倍数用于与下一个数字进行计算最大公倍数
idx += 1
return f
#累加法
def solution(*args) -> int:
if args == ():
return
if len(args) == 1:
return args[0]
mul = 1
for i in args:
mul *= i
j = max(args)
while max(args) <= j <= mul:
flag = True
for i in args:
if j % i != 0:
flag = False
if flag == True:
return j
j += 1
assert solution(2, 3, 4) == 12
assert solution(2, 5) == 10
assert solution(9) == 9
assert solution(3) == 3
优化了下，CodeWars过了。 
def solution(*args) -> int:
# your code here
len_args = len(args)
if 0 in args:
return 0
elif len_args == 0:
return 1
elif len_args == 1:
return args[0]
else:
greater = max(args)
nums = 0
while nums < len_args:
for i in args:
if greater % i == 0:
nums += 1
else:
greater += 1
nums = 0
break
return greater
assert solution(0, 1, 0) == 0
assert solution(2, 3, 4) == 12
assert solution(2, 5) == 10
assert solution(2, 4) == 4
assert solution(9) == 9
assert solution(1, 2, 3, 4, 5) == 60
assert solution(2, 4, 27, 8, 3, 9, 24, 216, 288, 51288) == 1846368
assert solution(2, 4, 27, 8, 3, 9, 24, 216, 288, 51288, 0) == 0
assert solution() == 1
# 辗转相减法求最大公因数
def solution(*args) -> int:
if len(args) == 0:
return -1
new_list = []
for i, i_value in enumerate(args):
for j_value in args[i + 1:]:
tmp = 1
mi = min(i_value, j_value)
ma = max(i_value, j_value)
while tmp != 0:
tmp = ma - mi
if tmp > mi:
ma = tmp
else:
ma = mi
mi = tmp
new_list.append(ma)
mu, p = 1, 1
for i in new_list:
p *= i
for i in args:
mu *= i
return mu / p
#上边的有点绕了，再简化
def solution(*args) -> int:
if len(args) == 0:
return -1
new_list = []
for i, i_value in enumerate(args):
for j_value in args[i + 1:]:
tmp = i_value
while tmp != j_value:
if tmp > j_value:
tmp = tmp - j_value
else:
j_value = j_value -tmp
new_list.append(j_value)
mu, p = 1, 1
for i in new_list:
p *= i
for i in args:
mu *= i
return mu / p
assert solution(2, 3, 4) == 12
assert solution(2, 5) == 10
assert solution(9) == 9
assert solution(3) == 3
assert solution() == -1
def solution(*args):
li=sorted(args)
boud = 1
while 1:
for i in li[1:]:
if (li[0]*boud)%i!=0:
break
else:
return li[0] * boud
boud+=1def solution(*args):
mul, mul_param = 1, 1
args = sorted(args)
for j, v in enumerate(args):
mul *= v
for m in args[j + 1:]:
n = v
while True:
r = m % n
if r == 0:
mul_param *= n
break
else:
m = n
n = r
return mul / mul_param
assert solution(2, 3, 4) == 12
assert solution(2, 5) == 10
assert solution(9) == 9
assert solution(3) == 3
def solution(*args)-> int:
# your code here
ls = [i for i in args]
if 0 in args:
return 0
elif ls == 0:
return 1
elif len(ls) == 1:
return ls[0]
elif len(ls) == 2:
return ls[0]*ls[1]
else:
num = max(ls)
a = 0
while True:
a+=1
count_num = a*num
l = 0
for i in ls:
if count_num % i ==0:
l+=1
if l ==len(ls):
return count_num
if __name__ == '__main__':
assert solution(2,3,4) == 12
assert solution(2,5) == 10
assert solution(9) == 9
assert solution(2, 4, 27, 8, 3, 9, 24, 216, 288, 51288) == 1846368
def solution(*args)-> int:
a = max([*args])
while True:
b=[a%i==0 for i in args]
if set(b)=={True}:
return a
a = a + 1
def solution(*args):
# your code here
# 转换为可变列表
l = [i for i in args]
# 输入为空数据，则返回None
if l == []:
return None
# 输入一个正整数，则直接返回该数据本身
if len(l) == 1:
return l[0]
# 输入正整数数量大于等于2时，每次取出来两个取最小公倍数，并把取出来的数据从列表删除，同时把取出来两个数的最小公倍数输入列表中
while len(l) >= 2:
n, m = l[0], l[1]
l.pop(0), l.pop(0)
s = []
tmp = 1
j = 2
while True:
if n % j == 0 and m % j == 0:
s.append(j)
n, m = n // j, m // j
continue
if j > n or j > m:
break
j += 1
for k in s:
tmp *= k
tmp *= n*m
l.append(tmp)
return l[0]
if len(args) == 0:
return 1
if 0 in args:
return 0
if len(args) == 1:
return args[0]
init_v = args[0]
for i in range(1,len(args)):
vx=max_v = max(args[i],init_v)
vn=min_v = min(args[i],init_v)
while max_v % min_v != 0:
max_v, min_v = min_v, max_v % min_v
init_v = vx * vn // min_v
return init_v
套用了思想 
def solution(*args)-> int:
max_num = max(args)
while True:
for num in args:
#将最大数分别对列表中的元素取余
if max_num % num != 0:
max_num += 1
continue
if num == args[-1]:
return max_num
assert solution(2,3,4) == 12
assert solution(2,5) == 10
assert solution(9) == 9
assert solution(2,4,27,8,3,9,24,216,288,51288) == 1846368
def solution(*args) -> int:
res = 1
while res >= 0:
l = 0
for i in args:
if res % int(i) != 0:
break
else:
l += 1
if l == len(args):
break
res += 1
return res
assert solution(2, 3, 4) == 12
assert solution(2, 5) == 10
assert solution(9) == 9
assert solution(2, 4, 27, 8, 3, 9, 24, 216, 288, 51288) == 1846368