I want to define a recursive function to merge two sorted lists (these two lists are sorted) and return a new list containing all the values in both argument lists with a incre
Just a simpler version:
def combine(a, b):
if a and b:
if a[0] > b[0]:
a, b = b, a
return [a[0]] + combine(a[1:], b)
return a + b
Test:
>>> combine([1,3,6,8], [2,4,5,7])
[1, 2, 3, 4, 5, 6, 7, 8]
Here are some alternatives:
The smarter way to do this is to use merge function from the heapq module:
from heapq import merge
list(merge(a,b))
Test:
>>> a = [1,2,3,4,7,9]
>>> b = [5,6,7,8,12]
>>> [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 12]
And without using recursion:
def combine(a:list, b:list):
alist = []
i,j = 0,0
while i < len(a) and j < len(b):
if a[i] < b[j]:
alist.append(a[i])
i+=1
else:
alist.append(b[j])
j+=1
while i < len(a):
alist.append(a[i])
i+=1
while j < len(b):
alist.append(b[j])
j+=1
return alist
Test:
>>> a = [1,2,3,4,7,9]
>>> b = [5,6,7,8,12]
>>> [1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 12]
Instead of return, you should add it to the alist as like below.
def combine(a, b):
alist = []
if a == [] and b == []:
return alist
if a != [] and b == []:
return alist + a
if a == [] and b != []:
return alist + b
if a != [] and b != []:
if a[0] <= b[0]:
alist.append(a[0])
alist = alist + combine(a[1:], b)
if a[0] > b[0]:
alist.append(b[0])
alist = alist + combine(a, b[1:])
return alist
def combine(a,b):
if not a and not b: return []
if not a: return [b[0]] + combine(a, b[1:])
if not b: return [a[0]] + combine(a[1:], b)
if a[0] > b[0]:
return [b[0]] + combine(a, b[1:])
return [a[0]] + combine(a[1:], b)
Your test case:
In [2]: a = [1,2,3,4]
In [3]: b = [5,6,7,8]
In [4]: combine(a,b)
Out[4]: [1, 2, 3, 4, 5, 6, 7, 8]
Another test case:
In [24]: a
Out[24]: [1, 2, 3, 8, 9]
In [25]: b
Out[25]: [1, 3, 5, 6, 7]
In [26]: combine(a,b)
Out[26]: [1, 1, 2, 3, 3, 5, 6, 7, 8, 9]