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This question was recently asked by Google.

Given two sorted linked lists, merge them in order.

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 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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TechLead and Joma

"TechLead" is Patrick Shyu - ex-Google tech lead, ex-Facebook Staff Software Engineer, multi-millionaire app entrepreneur, digital nomad, and software engineer. He has conducted over 100 interviews at Google.

"Joma" is Jonathan Ma - he has done internships and full-time at Citadel, Linkedin, Facebook and Microsoft. He has worked as a data scientist, software engineer and PM. He is currently working as a software engineer in FANG.