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Elso hazi #24

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2358111
list1
szellboglarka Mar 18, 2018
7ef3e4c
list2
szellboglarka Mar 18, 2018
48eeba0
string1
szellboglarka Mar 18, 2018
5d35389
string2
szellboglarka Mar 19, 2018
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20 changes: 14 additions & 6 deletions basic/list1.py
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Original file line number Diff line number Diff line change
Expand Up @@ -21,8 +21,11 @@
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
return
items = 0
for i in words:
if (len(i) > 1) and (i[0] == i[-1]):
items = items + 1
return items


# B. front_x
Expand All @@ -33,8 +36,14 @@ def match_ends(words):
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
return
beginsWithX= list()
notBeginsWithX = list()
for i in words:
if i[0] == 'x':
beginsWithX.append(i)
else:
notBeginsWithX.append(i)
return sorted(beginsWithX) + sorted(notBeginsWithX)


# C. sort_last
Expand All @@ -44,8 +53,7 @@ def front_x(words):
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
# +++your code here+++
return
return sorted(tuples,key=lambda x: x[1])


# Simple provided test() function used in main() to print
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26 changes: 22 additions & 4 deletions basic/list2.py
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Expand Up @@ -13,17 +13,35 @@
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
return
numsWithoutRepeating = list()
for i in nums:
if i not in numsWithoutRepeating:
numsWithoutRepeating.append(i)
return numsWithoutRepeating
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@szabadkai szabadkai Mar 22, 2018

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Pont azt csinalja amit iger a valtozo neve. sajnos ez nem az amit a feladat ker.
probald ki egy [1,2,3,1] teszt listaval.



# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
return
mergedList= list()
i =0
j=0
while (i < len(list1)) and (j < len (list2) ):
if list1[i] <= list2[j]:
mergedList.append(list1[i])
i= i+1
elif list1[i] > list2[j]:
mergedList.append(list2[j])
j=j+1
while (i < len(list1)):
mergedList.append(list1[i])
i= i+1
while (j < len(list2)):
mergedList.append(list2[j])
j = j + 1
return mergedList
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@szabadkai szabadkai Mar 22, 2018

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szuper 👍



# Note: the solution above is kind of cute, but unforunately list.pop(0)
Expand Down
28 changes: 18 additions & 10 deletions basic/string1.py
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Original file line number Diff line number Diff line change
Expand Up @@ -24,8 +24,10 @@
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
return
if count > 9:
return "Number of donuts: many"
else:
return "Number of donuts: " + str(count)


# B. both_ends
Expand All @@ -34,11 +36,15 @@ def donuts(count):
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
return
if len(s) < 2:
s = ''
return s
else:
return s[0] + s[1] + s[-2] + s[-1]

# C. fix_start


# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
Expand All @@ -48,8 +54,9 @@ def both_ends(s):
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
return
listOfChars = list(s.replace(s[0], '*'))
listOfChars[0] = s[0]
return ''.join(listOfChars)


# D. MixUp
Expand All @@ -60,14 +67,15 @@ def fix_start(s):
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
return
aSwapped = b[:2] + a[2:]
bSwapped = a[:2] + b[2:]
return aSwapped + ' ' + bSwapped


# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
if got== expected:
prefix = ' OK '
else:
prefix = ' X '
Expand Down
39 changes: 32 additions & 7 deletions basic/string2.py
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Original file line number Diff line number Diff line change
Expand Up @@ -16,8 +16,12 @@
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
return
if len(s) >= 3:
if s[-3:] == "ing":
s = s + "ly"
else:
s = s + "ing"
return s


# E. not_bad
Expand All @@ -29,9 +33,11 @@ def verbing(s):
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
return

index = s.find("not")
index2 = s.find("bad")
if index > 0 and index2 > 0 and index < index2:
s = s[:index] + "good" + s[index2 + 3:]
return s

# F. front_back
# Consider dividing a string into two halves.
Expand All @@ -41,8 +47,27 @@ def not_bad(s):
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
return
la = len(a)
lb = len(b)
afront = ""
bfront = ""
aback = ""
bback = ""
if divmod(la, 2) == 0:
afront = a[:la/2]
aback = a[(la/2) + 2:]
else:
afront = a[:(int)((la+1) / 2)]
aback = a[(int)((la-1) / 2):]

if divmod(lb, 2) == 0:
bfront = b[:lb/2]
bback = b[(lb/2) + 2:]
else:
bfront = b[:(int)((lb+1) / 2)]
bback = b[(int)((lb-1) / 2):]

return afront + bfront + aback + bback


# Simple provided test() function used in main() to print
Expand Down