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one day at a time.    ## Sample Coding Problem

Given two sorted linked lists, merge them in order.
 ```def merge(self, list1, list2): # Fill this in. # Definition for a Linked List. class ListNode(object): def __init__(self, x): self.val = x self.next = None ``` This can be done recursively and iteratively. See if you can get both solutions.

This problem can be solved recursively or iteratively. We traverse the two linked lists in parallel, advancing both pointers simultaneously. If the first list's value is smaller we advance that one, otherwise we advance the second list. If either list is shorter, then we take values from the longer list.

The time complexity is linear O(n) since both lists are traversed just once. The space complexity of the recursive algorithm is linear O(n), since it builds up a recursive stack that may be as deep as the length of both lists. The space complexity of the iterative solution is constant O(1), since only a few variables are used.

```# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None

class Solution:
def mergeTwoLists(self, l1, l2):
if l1 is None:
return l2
elif l2 is None:
return l1
elif l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2

def mergeTwoListsIterative(self, l1, l2):
current = None
root = None
while True:
if l1 is None:
nextNode = l2
elif l2 is None:
nextNode = l1
elif l1.val < l2.val:
nextNode = l1
else:
nextNode = l2

if nextNode == l1:
l1 = l1.next if l1 else None
if nextNode == l2:
l2 = l2.next if l2 else None

if nextNode is None:
break
if not current:
current = nextNode
root = current
else:
current.next = nextNode
current = nextNode
return root

# Test program
a = ListNode(1)
a.next = ListNode(3)
a.next.next = ListNode(5)

b = ListNode(2)
b.next = ListNode(4)
b.next.next = ListNode(6)

c = Solution().mergeTwoListsIterative(a, b)
while c:
print c.val
c = c.next
```

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