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Grind 75.3 Merge Two Sorted Lists

Solved at: 2022-07-13

Question

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Solution

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
start = []
while True:
if not list1 and not list2:
break
if not list1:
start.append(list2)
list2 = list2.next
continue
if not list2:
start.append(list1)
list1 = list1.next
continue
if list1.val <= list2.val:
start.append(list1)
list1 = list1.next
continue
else:
start.append(list2)
list2 = list2.next
continue
prev = None
for el in start:
if prev:
prev.next = el
prev = el
if not start:
return None
else:
return start[0]


I misunderstood the question, thinking list1 is type list instead of Node.

  • Time Complexity: O(n)O(n)
  • Space Complexity: O(n)O(n)

Results

Runtime

  • 33 ms, faster than 98.24% of Python3 online submissions for Merge Two Sorted Lists.

Memory Usage

  • 14 MB, less than 32.33% of Python3 online submissions for Merge Two Sorted Lists.

Other Answers Online

class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
cur = dummy = ListNode()
while list1 and list2:
if list1.val < list2.val:
cur.next = list1
list1, cur = list1.next, list1
else:
cur.next = list2
list2, cur = list2.next, list2

if list1 or list2:
cur.next = list1 if list1 else list2

return dummy.next
  • Time Complexity: O(n)O(n)
  • Space Complexity: O(1)O(1)

The above solution looks a lot cleaner and Pythonic.