A Recursive Function To Sort A List Of Ints
I want to define a recursive function can sort any list of ints: def sort_l(l): if l==[]: return [] else: if len(l)==1: return [l[-1]] e
Solution 1:
The quick sort is recursive and easy to implement in Python:
defquick_sort(l):
iflen(l) <= 1:
return l
else:
return quick_sort([e for e in l[1:] if e <= l[0]]) + [l[0]] +\
quick_sort([e for e in l[1:] if e > l[0]])
will give:
>>> quick_sort([3, 1, 2, 4, 7, 5, 6, 9, 8])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Solution 2:
For this you would want to use merge sort. Essentially in a merge sort you recursively split the list in half until you have single elements and than build it back up in the correct order. merge sort on has a complexity of O(n log(n))
and is an extremely stable sorting method.
Here are some good in depth explanations and visuals for merge sorting:
Solution 3:
def quicksort(lst):
"Quicksort over a list-like sequence"iflen(lst) == 0:
returnlstpivot= lst[0]
pivots = [x for x in lst ifx== pivot]
small = quicksort([x for x in lst if x < pivot])
large = quicksort([x for x in lst if x > pivot])
return small + pivots + large
Above is a more readable recursive implementation of Quick Sort Algorithm. Above piece of code is from book Functional programing in python by O'REILLY. Above function will produce.
list=[9,8,7,6,5,4]
quicksort(list)
>>[4,5,6,7,8,9]
Solution 4:
def sort(array, index = 0, bigNumber = 0):
if len(array) == index:
return array
elif bigNumber > array[index]:
array[index - 1] = array[index]
array[index] = bigNumber
bigNumber = array[0]
index = 0else:
bigNumber = array[index]
returnsort(array, (index + 1), bigNumber)
Solution 5:
#sort an int list using recursion
globalarrayarray=[5,3,8,4,2,6,1]
def sort1(array:[])->[]:
if len(array)==1:
return
temp=array[-1]
array.pop()
sort1(array)
sort2(array,temp)
def sort2(array:[],temp):
if len(array)==0or temp>=array[-1]:
array.append(temp)
return
a=array[-1]
array.pop()
sort2(array,temp)
array.append(a)
sort1(array)
print(array)
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