标签：The not returns oob func pre hang 多少 窗户
Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result.
It should remove all values from list a, which are present in list b.
array_diff([1,2],[1]) == [2]If a value is present in b, all of its occurrences must be removed from the other:
array_diff([1,2,2,2,3],[2]) == [1,3]
def array_diff(a, b): c = [] for i in a: if i not in b: c.append(i) return c or def array_diff(a, b): return [i for i in a if i not in b]
2.
Your job is to write a function which increments a string, to create a new string.
Examples:
foo -> foo1
foobar23 -> foobar24
foo0042 -> foo0043
foo9 -> foo10
foo099 -> foo100
不会
def increment_string(strng): head = strng.rstrip(‘0123456789‘) tail = strng[len(head):] print(head,tail) if tail == "": return strng + "1" return head + str(int(tail) + 1).zfill(len(tail))
3. 输入n 输出xy 是x^y == n
import math def isPP(n): i=2 while i < n: if i ** round(math.log(n,i)) == n: return [i,round(math.log(n,i))] else: i+=1 else: return None
4. h bounce window 输出 从窗户可以看到多少次球
h： 扔球的高度， 应大于0
bounce: 球重新弹起来的高度是原来的多少 （0，1)
window ： 看球的高度
def bouncingBall(h, bounce, window): if h >0 and bounce >0 and bounce<1 and window < h: a = 1 while h * bounce > window: a += 2 h = h*bounce else: return a else: return -1
5. 钞票25 50 100， 门票一张25 输入一个列表 ，看看能不能找钱给顾客
test.assert_equals(tickets([25, 25, 50]), "YES")
test.assert_equals(tickets([25, 100]), "NO")
def tickets(people): till = {100.0:0, 50.0:0, 25.0:0} for paid in people: till[paid] += 1 change = paid-25.0 for bill in (50,25): while (bill <= change and till[bill] > 0): till[bill] -= 1 change -= bill if change != 0: return ‘NO‘ return ‘YES‘
标签：The not returns oob func pre hang 多少 窗户
原文地址：https://www.cnblogs.com/adelinebao/p/13200926.html
