【每日一题0110】手串计数

兴趣爱好交流 每日一题
1

:woman_mage: 已知一个手串是由红色和蓝色珠子搭配构成的，并且每两个蓝色珠子之间夹着连续的2个红色珠子。
image

请编写一个函数，输入一个蓝色珠子的数量num，请求出红色珠子的个数。如果输入的数字小于2则返回0。

示例：
输入：1，输出：0。
输入：5，输出：8。

题目难度：简单
题目来源：CodeWars：Simple beads count

def solution(num: int) -> int:
    # your code

assert solution(1) == 0
assert solution(3) == 4
assert solution(5) == 8
2 
def solution(num: int) -> int:
    # your code
    if num < 2:
        return 0
    red = (num-1) * 2
    return red


assert solution(1) == 0
assert solution(3) == 4
assert solution(5) == 8
1 个赞
3 
def solution(num: int) -> int:
    if num < 2:
        return 0
    return (num - 1) * 2
1 个赞
4 
def solution(num: int) -> int:
    return 0 if num < 2 else (num-1)*2
5 
def solution0110(num: int) -> int:
    # 已知一个手串是由红色和蓝色珠子搭配构成的，并且每两个蓝色珠子之间夹着连续的2个红色珠子
    # 请编写一个函数，输入一个蓝色珠子的数量num，请求出红色珠子的个数。如果输入的数字小于2则返回0

    return (num-1)*2 if num >=2 else 0
6 
def solution16(num:int):
    if num<2:
        return 0
    else:
        return (num-1)*2
assert solution16(1) == 0
assert solution16(3) == 4
assert solution16(5) == 8
7 
def solution(num: int) -> int:
    return (num - 1) * 2 if num > 1 else 0
8 
def solution(num: int) -> int:
    return 0 if num<2 else (num-1)*2
9

def solution(num: int) → int:
return （num-1）*2

10 
def solution(num: int) -> int:
    return 2*(num-1) if num >1 else 0

assert solution(1) == 0
assert solution(3) == 4
assert solution(5) == 8
11 
def solution(num: int) -> int:
    return 0 if num < 2 else (num - 1) * 2


assert solution(1) == 0
assert solution(3) == 4
assert solution(5) == 8
12 
def solution(num: int) -> int:
    return (num - 1) * 2 if num > 0 else 0


assert solution(1) == 0
assert solution(3) == 4
assert solution(5) == 8
13 
def solution(num: int) -> int:
    return 0 if num<2 else (num-1)*2

assert solution(1) == 0
assert solution(3) == 4
assert solution(5) == 8