目录
1.判断电话号码算法挑战
如果传入字符串是一个有效的美国电话号码，则返回 true.
用户可以在表单中填入一个任意有效美国电话号码. 下面是一些有效号码的例子(还有下面测试时用到的一些变体写法):
在本节中你会看见如 800-692-7753 or 8oo-six427676;laskdjf这样的字符串. 你的任务就是验证前面给出的字符串是否是有效的美国电话号码. 区号是必须有的. 如果字符串中给出了国家代码, 你必须验证其是 1.如果号码有效就返回 true ; 否则返回 false.
def telephoneCheck(s):
i = 0
for x in s:
if x.isdigit(): i += 1
if "(" in s or ")" in s:
if "(" not in s or ")" not in s:
return False
if s[0] == "(" and s[-1] == ")": return False
if i == 10:
return True
elif i == 11:
return True if s[0] == "" else False
else:
return False print(telephoneCheck("1 555-555-5555")) # 应该返回 true.
print(telephoneCheck("1 (555) 555-5555")) # 应该返回 true.
print(telephoneCheck("555-555-5555")) # 应该返回 true.
print(telephoneCheck("(555)555-5555")) # 应该返回 true.
print(telephoneCheck("1(555)555-5555")) # 应该返回 true.
print(telephoneCheck("1 555)555-5555")) # 应该返回 false.
print(telephoneCheck("123**&!!asdf#")) # 应该返回 false.
print(telephoneCheck("(6505552368)")) # 应该返回 false
print(telephoneCheck("(275)76227382")) # 应该返回 false.
2.集合交集算法挑战
创建一个函数，接受两个或多个数组，返回所给数组的 对等差分(symmetric difference)(△ or ⊕)数组.
给出两个集合 (如集合 A = {1, 2, 3} 和集合 B = {2, 3, 4}), 而数学术语 "对等差分" 的集合就是指由所有只在两个集合其中之一的元素组成的集合(A △ B = C = {1, 4}). 对于传入的额外集合 (如 D = {2, 3}), 你应该安装前面原则求前两个集合的结果与新集合的对等差分集合 (C △ D = {1, 4} △ {2, 3} = {1, 2, 3, 4}).
sym([1, 2, 5], [2, 3, 5], [3, 4, 5]) 应该返回 [1, 4, 5]
import functools def sym(*args):
def diff(lst1, lst2):
lst = []
for i in lst1:
if i not in lst2:
lst.append(i)
for i in lst2:
if i not in lst1:
lst.append(i)
return list(set(lst)) lst = functools.reduce(diff, args)
return lst print(sym([1, 2, 3], [5, 2, 1, 4])) # 应该返回 [3, 4, 5].
print(sym([1, 2, 5], [2, 3, 5], [3, 4, 5])) # 应该返回 [1, 4, 5]
print(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])) # 应该返回 [1, 4, 5].
print(sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])) # 应该返回 [2, 3, 4, 6, 7].
print(sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])) # 应该返回 [1, 2, 4, 5, 6, 7, 8, 9].
3.收银系统算法挑战
设计一个收银程序 checkCashRegister() ，其把购买价格(price)作为第一个参数 , 付款金额 (cash)作为第二个参数, 和收银机中零钱 (cid) 作为第三个参数.
cid 是一个二维数组，存着当前可用的找零.
当收银机中的钱不够找零时返回字符串 "Insufficient Funds". 如果正好则返回字符串 "Closed".
否者, 返回应找回的零钱列表,且由大到小存在二维数组中.
from math import floor def checkCashRegister(price, cash, cid):
num = round(cash - price, 2) # 浮点数精度问题，用round函数精确到分位即可得出金额准确数值
dct = {"PENNY": 0.01, "NICKEL": 0.05, "DIME": 0.1, "QUARTER": 0.25, "ONE": 1, "FIVE": 5, "TEN": 10, "TWENTY": 20,
"ONE HUNDRED": 100}
dct_rev = {v: k for k, v in dct.items()}
# 面额作为键，数量作为值，去掉0和大于需找零金额的键，得到dct_aftdel
dct_aftdel = {dct[i[0]]: round(i[1] / dct[i[0]]) for i in cid if i[1] and dct[i[0]] <= num}
sum_all = sum([i[1] for i in cid if dct[i[0]] in dct_aftdel])
if sum_all < num:
return "Insufficient Funds"
elif sum_all == num:
return "Closed"
else:
print(f"要找{num}元")
print(f"从这些钱里面找：{dct_aftdel}")
print(f"这些钱里面一共有{sum_all}元")
dct_return = {}
for i in sorted(dct_aftdel, reverse=True):
if i <= num:
if floor(num / i) <= dct_aftdel[i]: # 例如：要找他90，你有5张20，只需找他4张，如果只有4张，则4张都找给他
dct_return[i] = floor(num / i) # 向下取整
print(f"找了{floor(num / i)}张{i}块的，还剩", end="")
num -= i * floor(num / i)
print(f"{round(num, 2)}")
num = round(num, 2) # num_temp为浮点型，自减完之后需要分位取整 else:
dct_return[i] = dct_aftdel[i]
num -= i * dct_aftdel[i]
print(f"找了{dct_aftdel[i]}张{i}块的，还剩{round(num, 2)}")
num = round(num, 2)
lst_return = [[dct_rev[k], k * v] for k, v in dct_return.items()]
return lst_return print(checkCashRegister(19.50, 20.00,
[["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00],
["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]]))
# 应该返回 [["QUARTER", 0.50]].
print(checkCashRegister(3.26, 100.00,
[["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00],
["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]]))
# 应该返回 [["TWENTY", 60.00], ["TEN", 20.00], ["FIVE", 15], ["ONE", 1], ["QUARTER", 0.50], ["DIME", 0.20], ["PENNY", 0.04]].
print(checkCashRegister(19.50, 20.00,
[["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0],
["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]))
# 应该返回 "Insufficient Funds".
print(checkCashRegister(19.50, 20.00,
[["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 1.00], ["FIVE", 0],
["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]))
# 应该返回 "Insufficient Funds".
print(checkCashRegister(19.50, 20.00,
[["PENNY", 0.50], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0],
["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]))
# 应该返回 "Closed".
4.库存更新算法挑战
依照一个存着新进货物的二维数组，更新存着现有库存(在 arr1 中)的二维数组. 如果货物已存在则更新数量 . 如果没有对应货物则把其加入到数组中，更新最新的数量. 返回当前的库存数组，且按货物名称的字母顺序排列.
updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]]) 应该返回 [[88, "Bowling Ball"], [2, "Dirty Sock"], [3, "Hair Pin"], [3, "Half-Eaten Apple"], [5, "Microphone"], [7, "Toothpaste"]].
def updateInventory(lst1, lst2):
lst_name = [i[1] for i in lst1] + [i[1] for i in lst2]
lst_quntity = [i[0] for i in lst1] + [i[0] for i in lst2]
dct = {}
# 字典去重名时计算数量
for i, j in zip(lst_name, lst_quntity):
if dct.get(i):
dct[i] = dct.get(i) + j
else:
dct[i] = j
lst = sorted([i for i in dct.items()]) # 按字母排序
lst = list(map(lambda x: list(reversed(x)), lst)) # 逆序
return lst print(updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]],
[[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]]))
# 应该返回 [[88, "Bowling Ball"], [2, "Dirty Sock"], [3, "Hair Pin"], [3, "Half-Eaten Apple"], [5, "Microphone"], [7, "Toothpaste"]]
print(updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]],
[]))
# 应该返回 [[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]].
print(updateInventory([], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7,
"Toothpaste"]]))
# 应该返回 [[67, "Bowling Ball"], [2, "Hair Pin"], [3, "Half-Eaten Apple"], [7, "Toothpaste"]].
print(updateInventory([[0, "Bowling Ball"], [0, "Dirty Sock"], [0, "Hair Pin"], [0, "Microphone"]],
[[1, "Hair Pin"], [1, "Half-Eaten Apple"], [1, "Bowling Ball"], [1,
"Toothpaste"]]))
# 应该返回 [[1, "Bowling Ball"], [0, "Dirty Sock"], [1, "Hair Pin"], [1, "Half-Eaten Apple"], [0, "Microphone"], [1, "Toothpaste"]].
5.排列组合去重算法挑战
把一个字符串中的字符重新排列生成新的字符串，返回新生成的字符串里没有连续重复字符的字符串个数.连续重复只以单个字符为准
例如, aab 应该返回 2 因为它总共有6中排列 (aab, aab, aba, aba, baa, baa), 但是只有两个 (aba and aba)没有连续重复的字符 (在本例中是 a).
permAlone("aaa") 应该返回 0.
permAlone("aabb") 应该返回 8.
import itertools def permAlone(s):
num = 0
for i in itertools.permutations(s, len(s)):
for x in range(len(i) - 1):
if i[x] == i[x + 1]:
break
else:
num += 1
return num print(permAlone("aab")) # 应该返回 2.
print(permAlone("aaa")) # 应该返回 0.
print(permAlone("abc")) # 应该返回 6.
print(permAlone("aabb")) # 应该返回 8.
print(permAlone("abcdefa")) # 应该返回 3600.
print(permAlone("abfdefa")) # 应该返回 2640.
print(permAlone("zzzzzzzz")) # 应该返回 0.
6.日期改写算法挑战
让日期区间更友好！
把常见的日期格式如：YYYY-MM-DD 转换成一种更易读的格式。
易读格式应该是用月份名称代替月份数字，用序数词代替数字来表示天 (1st 代替 1).
记住不要显示那些可以被推测出来的信息: 如果一个日期区间里结束日期与开始日期相差小于一年，则结束日期就不用写年份了。月份开始和结束日期如果在同一个月，则结束日期月份就不用写了。
另外, 如果开始日期年份是当前年份，且结束日期与开始日期小于一年，则开始日期的年份也不用写。
例如:
makeFriendlyDates(["2016-07-01", "2016-07-04"]) 应该返回 ["July 1st, 2016","4th"]
makeFriendlyDates(["2016-07-01", "2018-07-04"]) 应该返回 ["July 1st, 2016", "July 4th, 2018"].
makeFriendlyDates(["2016-12-01", "2017-02-03"]) should return ["December 1st, 2016","February 3rd"].
def makeFriendlyDates(lst):
dct1 = {1: "January", 2: "Februry", 3: "March", 4: "April", 5: "May", 6: "June", 7: "July", 8: "August",
9: "September", 10: "October", 11: "November", 12: "December"}
dct2 = {1: "1st", 2: "2nd", 3: "3rd", 4: "4th", 5: "5th", 6: "6th", 7: "7th", 8: "8th", 9: "9th", 10: "10th",
11: "11th", 12: "12th", 13: "13th", 14: "14th", 15: "15th", 16: "16th", 17: "17th", 18: "18th", 19: "19th",
20: "20th", 21: "21st", 22: "22nd", 23: "23rd", 24: "24th", 25: "25th", 26: "26th", 27: "27th", 28: "28th",
29: "29th", 30: "30th", 31: "31st"} date1 = lst[0].split("-")
date1 = list(map(int, date1))
date2 = lst[1].split("-")
date2 = list(map(int, date2)) if date1[0] == date2[0]: # 同年
if date1[1] == date2[1]:
if date1[2] == date2[2]:
return [f"{dct1[date1[1]]} {dct2[date1[2]]}", f" {date1[0]}"] # 同月同日
else:
return [f"{dct1[date1[1]]} {dct2[date1[2]]} {date1[0]}", f"{dct2[date2[2]]}"] # 同月不同日
else:
return [f"{dct1[date1[1]]} {dct2[date1[2]]}, {date1[0]}", f" {dct1[date2[1]]} {dct2[date2[2]]}"] # 不同月
else:
if date2[0] == date1[0] + 1 and (date2[1] < date1[1] or (date2[1] == date1[1] and date2[2] < date1[2])):
return [f"{dct1[date1[1]]} {dct2[date1[2]]}, {date1[0]}",
f" {dct1[date2[1]]} {dct2[date2[2]]}"] # 不同年相隔一年以内
else:
return [f"{dct1[date1[1]]} {dct2[date1[2]]}, {date1[0]}",
f" {dct1[date2[1]]} {dct2[date2[2]]}, {date2[0]}"] # 不同年相隔超过一年 print(makeFriendlyDates(["2016-07-01", "2016-07-04"]))
# 应该返回 ["July 1st, 2016","4th"]
print(makeFriendlyDates(["2016-07-01", "2018-07-04"]))
# 应该返回 ["July 1st, 2016", "July 4th, 2018"].
print(makeFriendlyDates(["2016-12-01", "2017-02-03"]))
# 应该返回 ["December 1st, 2016","February 3rd"].
print(makeFriendlyDates(["2016-12-01", "2018-02-03"]))
# 应该返回 ["December 1st, 2016","February 3rd, 2018"].
print(makeFriendlyDates(["2017-03-01", "2017-05-05"]))
# 应该返回 ["March 1st, 2017","May 5th"]
print(makeFriendlyDates(["2018-01-13", "2018-01-13"]))
# 应该返回 ["January 13th, 2018"].
print(makeFriendlyDates(["2022-09-05", "2023-09-04"]))
# 应该返回 ["September 5th, 2022","September 4th"].
print(makeFriendlyDates(["2022-09-05", "2023-09-05"]))
# 应该返回 ["September 5th, 2022","September 5th, 2023"]
7.类及对象构建算法挑战
用下面给定的方法构造一个对象.
方法有 getFirstName(), getLastName(), getFullName(), setFirstName(first), setLastName(last), and setFullName(firstAndLast).
所有有参数的方法只接受一个字符串参数.
所有的方法只与实体对象交互.
class Person:
def __init__(self, s):
self.fullname = s
self.firstname = s[:s.find(" ")]
self.lastname = s[s.find(" ") + 1:]
# self.fullname = self.firstname + " " + self.lastname def getFirstName(self):
print(f"FirstName是：{self.firstname}")
return self.firstname def getLastName(self):
print(f"LastName是：{self.lastname}")
return self.lastname def getFullName(self):
print(f"FullName是：{self.fullname}")
return self.fullname def setFirstName(self, s):
print(f"FirstName:{self.firstname}被修改为{s}")
self.firstname = s
self.fullname = s + " " + self.lastname def setLastName(self, s):
print(f"LastName:{self.lastname}被修改为{s}")
self.lastname = s
self.fullname = self.firstname + " " + s def setFullName(self, s):
print(f"FullName:{self.fullname}被修改为{s}")
self.fullname = s
self.firstname = s[:s.find(" ")]
self.lastname = s[s.find(" ") + 1:] bob = Person('Bob Ross')
bob.getFullName() bob.setFirstName("David")
bob.getFirstName() bob.setLastName("Alen")
bob.getLastName() bob.setFullName("Bruce Lee")
bob.getFullName()
8.轨道周期算法挑战
返回一个数组，其内容是把原数组中对应元素的平均海拔转换成其对应的轨道周期.
原数组中会包含格式化的对象内容，像这样 {name: 'name', avgAlt: avgAlt}.
至于轨道周期怎么求，戳这里 on wikipedia (不想看英文的话可以自行搜索以轨道高度计算轨道周期的公式).
求得的值应该是一个与其最接近的整数，轨道是以地球为基准的.
地球半径是 6367.4447 kilometers, 地球的GM值是 398600.4418, 圆周率为Math.PI
orbitalPeriod([{name : "sputnik", avgAlt : 35873.5553}]) 应该返回 [{name: "sputnik", orbitalPeriod: 86400}].
import math
def orbitalPeriod(lst):
r = 6367.4447
GM = 398600.4418
for dct in lst:
avgAlt = 2 * math.pi * math.sqrt(((dct["avgAlt"] + r) ** 3) / GM)
dct["orbitalPeriod"] = round(avgAlt)
del dct["avgAlt"]
return lst
print(orbitalPeriod([{"name": "sputnik", "avgAlt": 35873.5553}]))
print(orbitalPeriod(
[{"name": "iss", "avgAlt": 413.6}, {"name": "hubble", "avgAlt": 556.7}, {"name": "moon", "avgAlt": 378632.553}]))9.数据组合求值算法挑战
找到你的另一半
都说优秀的程序员擅长面向对象编程，但却经常找不到另一半，这是为什么呢？因为你总是把自己局限成为一个程序员，没有打开自己的思维。
这是一个社群的时代啊，在这里你应该找到与你有相同价值观但又互补的另一半。
譬如：你编程能力强，估值11分，如果以20分为最佳情侣来计算，你应该找一个设计能力强，估值为9分的女生。
那么当你遇到一个设计能力为9分的女生，千万别犹豫，大胆去表白。千万别以为后面的瓜比前面的甜哦。
举个例子：有一个能力数组[7,9,11,13,15]，按照最佳组合值为20来计算，只有7+13和9+11两种组合。而7在数组的索引为0，13在数组的索引为3，9在数组的索引为1，11在数组的索引为2。
所以我们说函数：pairwise([7,9,11,13,15],20) 的返回值应该是0+3+1+2的和，即6。
我们可以通过表格来更直观地查看数组中索引和值的关系：
| Index | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Value | 7 | 9 | 11 | 13 | 15 |
def pairwise(lst, n):
if not lst: return 0 # 空列表返回0
# 字典去重，元素作为键，重复值索引放列表作为值
dct = {j: [] for j in lst}
for d in dct:
dct[d] = [i for i, j in enumerate(lst) if j == d]
# print(dct)
# 当n为偶数，元素为n/2，且有奇数对时，删掉最后一个元素的索引值
if n % 2 == 0:
if lst.count(n / 2) % 2 == 1:
dct[n / 2].pop()
# 找另一半n-d，找不到则删键,找到了删右边多余的索引值
for d in dct.copy():
if n - d not in dct:
dct.pop(d)
else:
dct[d] = dct[d][:min(len(dct[d]), len(dct[n - d]))]
# print(dct)
# 加总所有字典值
result = sum([sum(i) for i in dct.values()])
return result print(pairwise([1, 4, 2, 3, 0, 5], 7)) # 应该返回 11.
print(pairwise([1, 3, 2, 4], 4)) # 应该返回 1.
print(pairwise([1, 1, 1], 2)) # 应该返回 1.
print(pairwise([0, 0, 0, 0, 1, 1, 2], 1)) # 应该返回 10.
print(pairwise([], 100)) # 应该返回 0.