【每日一题0920】最小公倍数

兴趣爱好交流 每日一题
1

:woman_mage:给定两个正整数 x 和 y ，请编写一个函数，找出它们最小公倍数。
示例：
输入：3,5，输出：15
输入：4,18，输出：36

题目难度：简单
题目来源：Udemy教程

def lcm(x: int, y: int) -> int:
    pass

assert lcm(3, 5) == 15
assert lcm(4, 16) == 16
assert lcm(4, 18) == 36
2 
 public Integer lcm(Integer x,Integer y){
        int max = Math.max(x, y);
        int min = Math.min(x, y);
        for (int i = 1; i <=min; i++) {
            if ((max*i)%min == 0){
                return max*i;
            }
        }
        throw new RuntimeException("no result");
    }
    @Test
    public void testLcm(){
        assert lcm(4, 16) == 16;
        assert lcm(3, 5) == 15;
        assert lcm(4, 18) == 36;
    }
3 
def gcd(x: int, y: int) -> int:
    # 辗转相除法求最大公约数
    if x < y:
        x, y = y, x
    while y != 0:
        x, y = y, x % y
    return x
  
def lcm(x: int, y: int) -> int:
    return x * y / gcd(x, y)
    
assert lcm(3, 5) == 15
assert lcm(4, 16) == 16
assert lcm(4, 18) == 36
4 
def lcm(x: int, y: int) -> int:
    max=x*y

    for i in range(1,max):
        if (i%x==0 and i%y==0):
            return i
    return max
5 
def lcm(x: int, y: int) -> int:
    max = x * y
    for i in range(1, max+1):
        if i % x == 0 and i % y == 0:
            return i
6 
def lcm(x: int, y: int) -> int:
    i = 1
    while(i):
        if i % x == 0 and i % y ==0:
            break
        else:
            i +=1
    return i

assert lcm(3, 5) == 15
assert lcm(4, 16) == 16
assert lcm(4, 18) == 36
7 
def lcm(x: int, y: int) -> int:
    max_mun = max(x, y)
    n = 1
    while True:
        common_divisor = max_mun * n
        if common_divisor % x == 0 and common_divisor % y == 0:
            return common_divisor
        n += 1


assert lcm(3, 5) == 15
assert lcm(4, 16) == 16
assert lcm(4, 18) == 36