21. Merge Two Sorted Lists

https://leetcode.com/problems/merge-two-sorted-lists/description/

 

이 문제는 풀긴 풀었다

# 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]:
        ans = ListNode(0)
        tmp = ans

        while list1 != None and list2 != None:
            if (list1.val <= list2.val):
                tmp.next = ListNode()
                tmp = tmp.next
                tmp.val = list1.val
                list1 = list1.next
            else:
                tmp.next = ListNode()
                tmp = tmp.next
                tmp.val = list2.val
                list2 = list2.next

        while list1:
            tmp.next = ListNode()
            tmp = tmp.next
            tmp.val = list1.val
            list1 = list1.next

        while list2:
            tmp.next = ListNode()
            tmp = tmp.next
            tmp.val = list2.val
            list2 = list2.next

        return ans.next

근데 https://ggodong.tistory.com/359 이 문제처럼 생각 덜 해가지고 코드가 개판이다.

 

리트코드는 코드의 품질도 중요하다고 하던데

 

또 생각안하고 풀면 뺨싸다구다

 

고친 후

 

# 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]:
        ans = ListNode(0)
        cur = ans

        while list1 != None and list2 != None:
            if (list1.val <= list2.val):
                cur.next = list1
                list1 = list1.next
            else:
                cur.next = list2
                list2 = list2.next
            
            cur = cur.next

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

        return ans.next