# 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.

## Solution​

# Definition for singly-linked list.# class ListNode:#     def __init__(self, val=0, next=None):#         self.val = val#         self.next = nextclass 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: $O(n)$
• Space Complexity: $O(n)$

## 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.

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: $O(n)$
• Space Complexity: $O(1)$