How to merge two sorted list in python in linear time(traverse only once)?

Question

list_a = [1,7,9,35,36,37]
list_b = [3,4,5,40]

Expected output:

list_merged = [1,3,4,5,7,9,35,36,37,40]

Condition : must traverse both list only once

I know of zip which works as below and does quite fit in my requirement.

>>> x = [1, 2]
>>> y = [1, 3, 4, 5, 6]
>>> zip(x,y)
[(1, 1), (2, 3)]
>>> zip(y, x)
[(1, 1), (3, 2)]

Answer 1

Here is what i tried myself. solution is not really short but is simple one..

def linear_merge(list1, list2):
  i = 0
  j = 0
  list_m = [] # resultant list

  while i < len(list1) and j < len(list2):
    if list1[i] <= list2[j]: #take element from list 1 
      list_m.append(list1[i])
      i += 1
    else: # or take element from list 2
      list_m.append(list2[j])
      j += 1

  if i <= len(list1) - 1: #append remaing elements from list 1 
    list_m.extend(list1[i:])
  elif j <= len(list2) - 1: #or append remaing elements from list 2
    list_m.extend(list2[j:])

  return list_m

This works at first hand, to me, seems the C way. Any more pythonic solution ?

Answer 2

You can traverse the two list with different index variables.

list_a = [1,7,9,35,36,37]
list_b = [3,4,5,40]
index_a = 0
index_b = 0
merged = []
while index_a<len(list_b) or index_b < len(list_b):
    if index_a<len(list_a) and list_a[index_a]<=list_b[index_b]:
        merged.append(list_a[index_a])
        index_a +=1
    elif index_b<len(list_b):
        merged.append(list_b[index_b])
        index_b +=1
print merged
# [1, 3, 4, 5, 7, 9, 35, 36, 37, 40]

Answer 3

The methodology is:

Keep two pointers(for the index of the lists) for each list and check the smaller element. When the smaller element is found move pointer next element for that list. Keep another(second list) pointer as it is. So you can traverse just one time for each list. If you implement above methodology, only thing you miss is that the biggest(last) element. So one more pointer is used to keep the last element.

Hope this helps:

list_a = [1,7,9,35,36,37] # also works for list_a = [1,7,9,35,36,45]
list_b = [3,4,5,40]

ptr_a = 0 # pointer for a
ptr_b = 0 # pointer for b
last = 0 # pointer for last element
list_merged = [] # merged result list

while ptr_a<len(list_a) and ptr_b<len(list_b):
    if list_a[ptr_a] < list_b[ptr_b]:
        list_merged.append(list_a[ptr_a])
        ptr_a = ptr_a + 1
        last = list_b[ptr_b] # assign last element

    else:
        list_merged.append(list_b[ptr_b])
        ptr_b = ptr_b + 1
        last = list_a[ptr_a] # assign last element

# at the end of the while loop the last element of 
# the 'merged list' has not added. Below line is to add it.
list_merged.append(last)
print list_merged

Answer 4

The easiest way to merge two lists is to use the '+' operator for the two

---

This is more of a pythonic way than the algorithmic way.

list1 = [1,2,3]
list2 = [4,5,6]

list3 = list1+list2 

#This merges both lists and puts them in list3

If you want a more algorithmic approach ....

use the below code :-

list3 = [ ]
for i in list1:
   list3.append(i);
for i in list2:
   list3.append(i) ;

Now list3 will have the merged result.

How ever I should warn you that , the first pythonic way is almost 3 times faster than the second way. (just tested it with timeit module)