W3C高級算法挑戰之python實現

時間 2019-11-17

標籤 w3c 高級算法挑戰 python 實現欄目 C&C++ 简体版

原文原文鏈接

最近在學python，網上很難找到對應的算法題網站，專業算法網站大部分都是國外的，以前在w3cschool看到有三個級別的Javascript腳本算法挑戰，嘗試用python實現，代碼量相對比較少，若是你有更好的解法，還請不吝賜教，初學python，但願和你們一塊兒日有所長。python

目錄git

1.判斷電話號碼算法挑戰程序員

1.判斷電話號碼算法挑戰

若是傳入字符串是一個有效的美國電話號碼，則返回 true.

用戶能夠在表單中填入一個任意有效美國電話號碼. 下面是一些有效號碼的例子(還有下面測試時用到的一些變體寫法):

555-555-5555 (555)555-5555 (555) 555-5555 555 555 5555 5555555555 1 555 555 5555

在本節中你會看見如 800-692-7753 or 8oo-six427676;laskdjf這樣的字符串. 你的任務就是驗證前面給出的字符串是不是有效的美國電話號碼. 區號是必須有的. 若是字符串中給出了國家代碼, 你必須驗證其是 1.若是號碼有效就返回 true ; 不然返回 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] == "1" else False
    else:
        return False


print(telephoneCheck("1 555-555-5555"))  # 應該返回 true.
print(telephoneCheck("1 (555) 555-5555"))  # 應該返回 true.
print(telephoneCheck("555-555-5555"))  # 應該返回 true.
print(telephoneCheck("(555)555-5555"))  # 應該返回 true.
print(telephoneCheck("1(555)555-5555"))  # 應該返回 true.
print(telephoneCheck("1 555)555-5555"))  # 應該返回 false.
print(telephoneCheck("123**&!!asdf#"))  # 應該返回 false.
print(telephoneCheck("(6505552368)"))  # 應該返回 false
print(telephoneCheck("(275)76227382"))  # 應該返回 false.

2.集合交集算法挑戰

建立一個函數，接受兩個或多個數組，返回所給數組的對等差分(symmetric difference)(△ or ⊕)數組.

給出兩個集合 (如集合 A = {1, 2, 3} 和集合 B = {2, 3, 4}), 而數學術語 "對等差分" 的集合就是指由全部只在兩個集合其中之一的元素組成的集合(A △ B = C = {1, 4}). 對於傳入的額外集合 (如 D = {2, 3}), 你應該安裝前面原則求前兩個集合的結果與新集合的對等差分集合 (C △ D = {1, 4} △ {2, 3} = {1, 2, 3, 4}).

sym([1, 2, 5], [2, 3, 5], [3, 4, 5]) 應該返回 [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]))  # 應該返回 [3, 4, 5].
print(sym([1, 2, 5], [2, 3, 5], [3, 4, 5]))  # 應該返回 [1, 4, 5]
print(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]))  # 應該返回 [1, 4, 5].
print(sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3]))  # 應該返回 [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]))  #  應該返回 [1, 2, 4, 5, 6, 7, 8, 9].

3.收銀系統算法挑戰

設計一個收銀程序 checkCashRegister() ，其把購買價格(price)做爲第一個參數 , 付款金額 (cash)做爲第二個參數, 和收銀機中零錢 (cid) 做爲第三個參數.

cid 是一個二維數組，存着當前可用的找零.

當收銀機中的錢不夠找零時返回字符串 "Insufficient Funds". 若是正好則返回字符串 "Closed".

否者, 返回應找回的零錢列表,且由大到小存在二維數組中.

數據轉換有點麻煩，整型和浮點型，字典與列表，解決的關鍵在於面額夠不夠找，找多少張的邏輯判斷，感受不是很完善，算法還能夠改進。

from math import floor


def checkCashRegister(price, cash, cid):
    num = round(cash - price, 2)  # 浮點數精度問題，用round函數精確到分位便可得出金額準確數值
    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()}
    # 面額做爲鍵，數量做爲值，去掉0和大於需找零金額的鍵，獲得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"從這些錢裏面找：{dct_aftdel}")
        print(f"這些錢裏面一共有{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張20，只需找他4張，若是隻有4張，則4張都找給他
                    dct_return[i] = floor(num / i)  # 向下取整
                    print(f"找了{floor(num / i)}張{i}塊的，還剩", end="")
                    num -= i * floor(num / i)
                    print(f"{round(num, 2)}")
                    num = round(num, 2)  # num_temp爲浮點型，自減完以後須要分位取整

                else:
                    dct_return[i] = dct_aftdel[i]
                    num -= i * dct_aftdel[i]
                    print(f"找了{dct_aftdel[i]}張{i}塊的，還剩{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]]))
# 應該返回 [["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]]))
# 應該返回 [["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]]))
# 應該返回 "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]]))
# 應該返回 "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]]))
# 應該返回 "Closed".

4.庫存更新算法挑戰

依照一個存着新進貨物的二維數組，更新存着現有庫存(在 arr1 中)的二維數組. 若是貨物已存在則更新數量 . 若是沒有對應貨物則把其加入到數組中，更新最新的數量. 返回當前的庫存數組，且按貨物名稱的字母順序排列.

updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]]) 應該返回 [[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 = {}
    # 字典去重名時計算數量
    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"]]))
 # 應該返回 [[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"]],
                      []))  
 # 應該返回 [[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"]]))  
 # 應該返回 [[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"]]))  
 # 應該返回 [[1, "Bowling Ball"], [0, "Dirty Sock"], [1, "Hair Pin"], [1, "Half-Eaten Apple"], [0, "Microphone"], [1, "Toothpaste"]].

5.排列組合去重算法挑戰

把一個字符串中的字符從新排列生成新的字符串，返回新生成的字符串裏沒有連續重複字符的字符串個數.連續重複只以單個字符爲準

例如, aab 應該返回 2 由於它總共有6中排列 (aab, aab, aba, aba, baa, baa), 可是隻有兩個 (aba and aba)沒有連續重複的字符 (在本例中是 a).

permAlone("aaa") 應該返回 0.

permAlone("aabb") 應該返回 8.

本身想沒作出來，偷個巧python內置排列組合函數，嘻嘻

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")) # 應該返回 2.
print(permAlone("aaa")) # 應該返回 0.
print(permAlone("abc")) # 應該返回 6.
print(permAlone("aabb")) # 應該返回 8.
print(permAlone("abcdefa")) # 應該返回 3600.
print(permAlone("abfdefa")) # 應該返回 2640.
print(permAlone("zzzzzzzz")) # 應該返回 0.

6.日期改寫算法挑戰

讓日期區間更友好！

把常見的日期格式如：YYYY-MM-DD 轉換成一種更易讀的格式。

易讀格式應該是用月份名稱代替月份數字，用序數詞代替數字來表示天 (1st 代替 1).

記住不要顯示那些能夠被推測出來的信息: 若是一個日期區間裏結束日期與開始日期相差小於一年，則結束日期就不用寫年份了。月份開始和結束日期若是在同一個月，則結束日期月份就不用寫了。

另外, 若是開始日期年份是當前年份，且結束日期與開始日期小於一年，則開始日期的年份也不用寫。

例如:

makeFriendlyDates(["2016-07-01", "2016-07-04"]) 應該返回 ["July 1st, 2016","4th"]

makeFriendlyDates(["2016-07-01", "2018-07-04"]) 應該返回 ["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]]}"]  # 同月不一樣日
        else:
            return [f"{dct1[date1[1]]} {dct2[date1[2]]}, {date1[0]}", f" {dct1[date2[1]]} {dct2[date2[2]]}"]  # 不一樣月
    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]]}"]  # 不一樣年相隔一年之內
        else:
            return [f"{dct1[date1[1]]} {dct2[date1[2]]}, {date1[0]}",
                    f" {dct1[date2[1]]} {dct2[date2[2]]}, {date2[0]}"]  # 不一樣年相隔超過一年


print(makeFriendlyDates(["2016-07-01", "2016-07-04"]))
# 應該返回 ["July 1st, 2016","4th"]
print(makeFriendlyDates(["2016-07-01", "2018-07-04"]))
# 應該返回 ["July 1st, 2016", "July 4th, 2018"].
print(makeFriendlyDates(["2016-12-01", "2017-02-03"]))
# 應該返回 ["December 1st, 2016","February 3rd"].
print(makeFriendlyDates(["2016-12-01", "2018-02-03"]))
# 應該返回 ["December 1st, 2016","February 3rd, 2018"].
print(makeFriendlyDates(["2017-03-01", "2017-05-05"]))
# 應該返回 ["March 1st, 2017","May 5th"]
print(makeFriendlyDates(["2018-01-13", "2018-01-13"]))
# 應該返回 ["January 13th, 2018"].
print(makeFriendlyDates(["2022-09-05", "2023-09-04"]))
# 應該返回 ["September 5th, 2022","September 4th"].
print(makeFriendlyDates(["2022-09-05", "2023-09-05"]))
# 應該返回 ["September 5th, 2022","September 5th, 2023"]

7.類及對象構建算法挑戰

用下面給定的方法構造一個對象.

方法有 getFirstName(), getLastName(), getFullName(), setFirstName(first), setLastName(last), and setFullName(firstAndLast).

全部有參數的方法只接受一個字符串參數.

全部的方法只與實體對象交互.

python基礎的構造類、魔法初始化方法、實例方法

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}被修改成{s}")
        self.firstname = s
        self.fullname = s + " " + self.lastname

    def setLastName(self, s):
        print(f"LastName:{self.lastname}被修改成{s}")
        self.lastname = s
        self.fullname = self.firstname + " " + s

    def setFullName(self, s):
        print(f"FullName:{self.fullname}被修改成{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.軌道週期算法挑戰

返回一個數組，其內容是把原數組中對應元素的平均海拔轉換成其對應的軌道週期.

原數組中會包含格式化的對象內容，像這樣 {name: 'name', avgAlt: avgAlt}.

至於軌道週期怎麼求，戳這裏 on wikipedia (不想看英文的話能夠自行搜索以軌道高度計算軌道週期的公式).

求得的值應該是一個與其最接近的整數，軌道是以地球爲基準的.

地球半徑是 6367.4447 kilometers, 地球的GM值是 398600.4418, 圓周率爲Math.PI

orbitalPeriod([{name : "sputnik", avgAlt : 35873.5553}]) 應該返回 [{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.數據組合求值算法挑戰

找到你的另外一半

都說優秀的程序員擅長面向對象編程，但卻常常找不到另外一半，這是爲何呢？由於你老是把本身侷限成爲一個程序員，沒有打開本身的思惟。

這是一個社羣的時代啊，在這裏你應該找到與你有相同價值觀但又互補的另外一半。

譬如：你編程能力強，估值11分，若是以20分爲最佳情侶來計算，你應該找一個設計能力強，估值爲9分的女生。

那麼當你遇到一個設計能力爲9分的女生，千萬別猶豫，大膽去表白。千萬別覺得後面的瓜比前面的甜哦。

舉個例子：有一個能力數組[7,9,11,13,15]，按照最佳組合值爲20來計算，只有7+13和9+11兩種組合。而7在數組的索引爲0，13在數組的索引爲3，9在數組的索引爲1，11在數組的索引爲2。

因此咱們說函數：pairwise([7,9,11,13,15],20) 的返回值應該是0+3+1+2的和，即6。

咱們能夠經過表格來更直觀地查看數組中索引和值的關係：

Index	0	1	2	3	4
Value	7	9	11	13	15

思路很重要，先寫出本身的處理邏輯，而後逐步分析邏輯中的數學方法，對應選擇容器和方法來作

def pairwise(lst, n):
    if not lst: return 0  # 空列表返回0
    # 字典去重，元素做爲鍵，重複值索引放列表做爲值
    dct = {j: [] for j in lst}
    for d in dct:
        dct[d] = [i for i, j in enumerate(lst) if j == d]
    # print(dct)
    # 當n爲偶數，元素爲n/2，且有奇數對時，刪掉最後一個元素的索引值
    if n % 2 == 0:
        if lst.count(n / 2) % 2 == 1:
            dct[n / 2].pop()
    # 找另外一半n-d，找不到則刪鍵,找到了刪右邊多餘的索引值
    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)
    # 加總全部字典值
    result = sum([sum(i) for i in dct.values()])
    return result


print(pairwise([1, 4, 2, 3, 0, 5], 7))  # 應該返回 11.
print(pairwise([1, 3, 2, 4], 4))  # 應該返回 1.
print(pairwise([1, 1, 1], 2))  # 應該返回 1.
print(pairwise([0, 0, 0, 0, 1, 1, 2], 1))  # 應該返回 10.
print(pairwise([], 100))  # 應該返回 0.