python练习园地(一)——Warmup篇

python练习题园地(一)——Warmup篇

最近找到一个网站:https://codingbat.com/python,这里是python和java初学者的园地,在这里你可以充分的发挥你在程序学习上的逻辑推理能力。一关一关的过,做你没做过或是曾做过后练习题,在学习的同时还可以熟悉一下英文,大多数都可以看懂。

Warmup-1:

Warmup-1 > sleep_in:

The parameter weekday is True if it is a weekday, and the parameter vacation is True if we are on vacation. We sleep in if it is not a weekday or we’re on vacation. Return True if we sleep in.
sleep_in(False, False) → True
sleep_in(True, False) → False
sleep_in(False, True) → True

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def sleep_in(weekday, vacation):
if not weekday and not vacation:
return True
elif weekday and not vacation:
return False
elif not weekday and vacation:
return True
else:
return True

Warmup-1 > monkey_trouble

We have two monkeys, a and b, and the parameters a_smile and b_smile indicate if each is smiling. We are in trouble if they are both smiling or if neither of them is smiling. Return True if we are in trouble.
monkey_trouble(True, True) → True
monkey_trouble(False, False) → True
monkey_trouble(True, False) → False

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def monkey_trouble(a_smile, b_smile):
if a_smile and b_smile:
return True
elif not a_smile and not b_smile:
return True
else:
return False

Warmup-1 > sum_double

Given two int values, return their sum. Unless the two values are the same, then return double their sum.
sum_double(1, 2) → 3
sum_double(3, 2) → 5
sum_double(2, 2) → 8

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def sum_double(a, b):
if a!=b:
return a+b
else:
return (a+b)*2

Warmup-1 > diff21

Given an int n, return the absolute difference between n and 21, except return double the absolute difference if n is over 21.
diff21(19) → 2
diff21(10) → 11
diff21(21) → 0

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def diff21(n):
if n<=21:
return 21-n
else:
return (n-21)*2

Warmup-1 > parrot_trouble

We have a loud talking parrot. The “hour” parameter is the current hour time in the range 0..23. We are in trouble if the parrot is talking and the hour is before 7 or after 20. Return True if we are in trouble.
parrot_trouble(True, 6) → True
parrot_trouble(True, 7) → False
parrot_trouble(False, 6) → False

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def parrot_trouble(talking, hour):
if talking and (hour<7 or hour>20):
return True
else:
return False

Warmup-1 > makes10

Given 2 ints, a and b, return True if one if them is 10 or if their sum is 10.
makes10(9, 10) → True
makes10(9, 9) → False
makes10(1, 9) → True

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def makes10(a, b):
if a==10 or b==10 or a+b==10:
return True
else:
return False

简洁版(官方版),仔细思量:

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def makes10(a, b):
return (a == 10 or b == 10 or a+b == 10)

Warmup-1 > near_hundred

Given an int n, return True if it is within 10 of 100 or 200. Note: abs(num) computes the absolute value of a number.
near_hundred(93) → True
near_hundred(90) → True
near_hundred(89) → False

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def near_hundred(n):
if abs(100-n)<=10 or abs(200-n)<=10 :
return True
else:
return False

官方版:

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def near_hundred(n):
return ((abs(100 - n) <= 10) or (abs(200 - n) <= 10))

Warmup-1 > pos_neg

Given 2 int values, return True if one is negative(负数) and one is positive(正数). Except if the parameter “negative” is True, then return True only if both are negative.
pos_neg(1, -1, False) → True
pos_neg(-1, 1, False) → True
pos_neg(-4, -5, True) → True

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def pos_neg(a, b, negative):
if ((a<0 and b>0) or(a>0 and b<0)) and not negative:
return True
elif negative and (a<0 and b<0) :
return True
else:
return False

官方版:

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def pos_neg(a, b, negative):
if negative:
return (a < 0 and b < 0)
else:
return ((a < 0 and b > 0) or (a > 0 and b < 0))

Warmup-1 > not_string

Given a string, return a new string where “not “ has been added to the front. However, if the string already begins with “not”, return the string unchanged. (给定一个字符串,返回一个新字符串,将“not ”添加到字符串开头。但是,如果字符串已经以“not”开头,则不加更改地返回字符串。)
not_string(‘candy’) → ‘not candy’
not_string(‘x’) → ‘not x’
not_string(‘not bad’) → ‘not bad’

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def not_string(str):
if 'not' in str[0:3]:
return str
else:
return "not "+str

官方版:

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def not_string(str):
if len(str) >= 3 and str[:3] == "not":
return str
return "not " + str
# str[:3] goes from the start of the string up to but not
# including index 3

Warmup-1 > missing_char

Given a non-empty string and an int n, return a new string where the char at index n has been removed. The value of n will be a valid index of a char in the original string (i.e. n will be in the range 0..len(str)-1 inclusive).
missing_char(‘kitten’, 1) → ‘ktten’
missing_char(‘kitten’, 0) → ‘itten’
missing_char(‘kitten’, 4) → ‘kittn’

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def missing_char(str, n):
return str[:n]+str[n+1:]

官方版:

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def missing_char(str, n):
front = str[:n] # up to but not including n
back = str[n+1:] # n+1 through end of string
return front + back

Warmup-1 > front_back

Given a string, return a new string where the first and last chars have been exchanged. (返回首尾字符交换的字符串)
front_back(‘code’) → ‘eodc’
front_back(‘a’) → ‘a’
front_back(‘ab’) → ‘ba’
以下代码本机测试通过(考虑字符串只有1,或0时):

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def front_back(str):
if len(str)<=1:
return str
else:
l=len(str)
first=str[0]
last=str[l-1]
return last+str[1:(l-1)]+first

官方版:

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def front_back(str):
if len(str) <= 1:
return str
mid = str[1:len(str)-1] # can be written as str[1:-1]
# last + mid + first
return str[len(str)-1] + mid + str[0]

Warmup-1 > front3

Given a string, we’ll say that the front is the first 3 chars of the string. If the string length is less than 3, the front is whatever is there. Return a new string which is 3 copies of the front.
front3(‘Java’) → ‘JavJavJav’
front3(‘Chocolate’) → ‘ChoChoCho’
front3(‘abc’) → ‘abcabcabc’

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def front3(str):
if len(str)<3:
return str*3
else:
return str[:3]*3

官方代码:

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def front3(str):
# Figure the end of the front
front_end = 3
if len(str) < front_end:
front_end = len(str)
front = str[:front_end]
return front + front + front

# Could omit the if logic, and write simply front = str[:3]
# since the slice is silent about out-of-bounds conditions.

Warmup-2

Warmup-2 > string_times

Given a string and a non-negative int n, return a larger string that is n copies of the original string. (给一个字符串和非负整数,返回整数倍的字符串)
string_times(‘Hi’, 2) → ‘HiHi’
string_times(‘Hi’, 3) → ‘HiHiHi’
string_times(‘Hi’, 1) → ‘Hi’

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def string_times(str, n):
return str*n

官方代码:

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def string_times(str, n):
result = ""
for i in range(n): # range(n) is [0, 1, 2, .... n-1]
result = result + str # could use += here
return result

Warmup-2 > front_times

Given a string and a non-negative int n, we’ll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front;(返回字符串的前3个字符的N个副本,小于3则返回3个)
front_times(‘Chocolate’, 2) → ‘ChoCho’
front_times(‘Chocolate’, 3) → ‘ChoChoCho’
front_times(‘Abc’, 3) → ‘AbcAbcAbc’

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def front_times(str, n):
if len(str)<3:
return str*n
else:
return str[:3]*n

官方代码:

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def front_times(str, n):
front_len = 3
if front_len > len(str):
front_len = len(str)
front = str[:front_len]

result = ""
for i in range(n):
result = result + front
return result

Warmup-2 > string_bits

Given a string, return a new string made of every other char starting with the first, so “Hello” yields “Hlo”.
string_bits(‘Hello’) → ‘Hlo’
string_bits(‘Hi’) → ‘H’
string_bits(‘Heeololeo’) → ‘Hello’

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def string_bits(str):
result=''
if len(str)<=1: #这两句多于了,有i%2==0足够表达
result=str
else:
for i in range(len(str)):
if i%2==0:
result=result+str[i]
return result

官方代码:

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def string_bits(str):
result = ""
# Many ways to do this. This uses the standard loop of i on every char,
# and inside the loop skips the odd index values.
for i in range(len(str)):
if i % 2 == 0:
result = result + str[i]
return result

Warmup-2 > string_splosion

Given a non-empty string like “Code” return a string like “CCoCodCode”.
string_splosion(‘Code’) → ‘CCoCodCode’
string_splosion(‘abc’) → ‘aababc’
string_splosion(‘ab’) → ‘aab’

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def string_splosion(str):
result=''
for i in range(len(str)):
if i==0: #这三句多余了,有str[:i+1]就足以包含,参考官方代码
result=result+str[i]
else:
result=result+str[:i+1]
return result

官方代码:

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def string_splosion(str):
result = ""
# On each iteration, add the substring of the chars 0..i
for i in range(len(str)):
result = result + str[:i+1]
return result

Warmup-2 > last2

Given a string, return the count of the number of times that a substring length 2 appears in the string and also as the last 2 chars of the string, so “hixxxhi” yields 1 (we won’t count the end substring). (返回字符串最后两个字符出现的次数,不计最后出现这一次)
last2(‘hixxhi’) → 1
last2(‘xaxxaxaxx’) → 1
last2(‘axxxaaxx’) → 2

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def last2(str):
count=0
str_len=len(str)
last=str[str_len-2:str_len]
for i in range(str_len):
if str[i:i+2]==last and i<str_len-2:
count+=1

return count

官方代码:

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def last2(str):
# Screen out too-short string case.
if len(str) < 2:
return 0

# last 2 chars, can be written as str[-2:]
last2 = str[len(str)-2:]
count = 0

# Check each substring length 2 starting at i
for i in range(len(str)-2):
sub = str[i:i+2]
if sub == last2:
count = count + 1

return count

Warmup-2 > array_count9

Given an array of ints, return the number of 9’s in the array.(返回整数数组中9出现的次数)
array_count9([1, 2, 9]) → 1
array_count9([1, 9, 9]) → 2
array_count9([1, 9, 9, 3, 9]) → 3

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def array_count9(nums):
count=0
for each in nums:
if each==9:
count+=1
return count

官方代码一致,唯变量不同:

Warmup-2 > array_front9

Given an array of ints, return True if one of the first 4 elements in the array is a 9. The array length may be less than 4. (返回整数数组中前4位数中出现9的布尔值)
array_front9([1, 2, 9, 3, 4]) → True
array_front9([1, 2, 3, 4, 9]) → False
array_front9([1, 2, 3, 4, 5]) → False

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def array_front9(nums):
return (9 in nums[:4])

官方代码:

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def array_front9(nums):
# First figure the end for the loop
end = len(nums)
if end > 4:
end = 4

for i in range(end): # loop over index [0, 1, 2, 3]
if nums[i] == 9:
return True
return False

Warmup-2 > array123

Given an array of ints, return True if the sequence of numbers 1, 2, 3 appears in the array somewhere.
array123([1, 1, 2, 3, 1]) → True
array123([1, 1, 2, 4, 1]) → False
array123([1, 1, 2, 1, 2, 3]) → True

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def array123(nums):
if len(nums)==0:
return False
for i in range(len(nums)):
if i+2<len(nums):
if nums[i]==1 and nums[i+1]==2 and nums[i+2]==3:
return True
else:
return False

官方代码:

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def array123(nums):
# Note: iterate with length-2, so can use i+1 and i+2 in the loop
for i in range(len(nums)-2):
if nums[i]==1 and nums[i+1]==2 and nums[i+2]==3:
return True
return False

Warmup-2 > string_match

Given 2 strings, a and b, return the number of the positions where they contain the same length 2 substring. So “xxcaazz” and “xxbaaz” yields 3, since the “xx”, “aa”, and “az” substrings appear in the same place in both strings. (返回两个字符串中在相同位置出现相同两个字符的次数)
string_match(‘xxcaazz’, ‘xxbaaz’) → 3
string_match(‘abc’, ‘abc’) → 2
string_match(‘abc’, ‘axc’) → 0

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def string_match(a, b):
if len(a)<=len(b): #以最短的字串长度为主
len_s=len(a)
else:
len_s=len(b)
count=0
for i in range(len_s-1):
if a[i]+a[i+1] == b[i]+b[i+1]:
count+=1
return count

官方代码:

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def string_match(a, b):
# Figure which string is shorter.
shorter = min(len(a), len(b))
count = 0

# Loop i over every substring starting spot.
# Use length-1 here, so can use char str[i+1] in the loop
for i in range(shorter-1):
a_sub = a[i:i+2]
b_sub = b[i:i+2]
if a_sub == b_sub:
count = count + 1

return count

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