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[chapse57] Week 4 #2745
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[chapse57] Week 4 #2745
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c474203
Add contains-duplicate solution
chapse57 9e53265
[chapse57] Week 1
chapse57 6895637
fix: add final newline
chapse57 afde7ad
[chapse57] Week 2
chapse57 888fb59
[chapse57] Week 2 - Group Anagrams
chapse57 9cb02e4
fix: add final newline
chapse57 b21959a
[chapse57] Week 3
chapse57 0b463ca
[chapse57] Week 3 - Number of 1 Bits
chapse57 e1da943
[chapse57] Week 4 - Merge Two Sorted Lists
chapse57 9f9ef0a
A
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,28 @@ | ||
| # Definition for singly-linked list. | ||
| # class ListNode(object): | ||
| # def __init__(self, val=0, next=None): | ||
| # self.val = val | ||
| # self.next = next | ||
| class Solution(object): | ||
| def mergeTwoLists(self, list1, list2): | ||
| """ | ||
| :type list1: Optional[ListNode] | ||
| :type list2: Optional[ListNode] | ||
| :rtype: Optional[ListNode] | ||
| """ | ||
| dummy = ListNode() | ||
| tail = dummy | ||
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| while list1 and list2: | ||
| if list1.val <= list2.val: | ||
| tail.next = list1 | ||
| list1 = list1.next | ||
| else: | ||
| tail.next = list2 | ||
| list2 = list2.next | ||
|
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| tail = tail.next | ||
| tail.next = list1 if list1 else list2 | ||
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| return dummy.next | ||
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 깔끔한 풀이네요! 다른 부분은 쉽게 이해가 됐는데 dummy 가 무슨 값을 가지는지 생각을 해야했던 거 같습니다. dummy_head 같은 약간은 의미를 드러내는 것이면 어떨까 생각해봤습니다. |
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🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 두 리스트를 순차적으로 비교하며 새 노드 없이 기존 노드를 재사용해 병합한다. 추가 포인터/더미 노드를 사용해 구현이 명확하다.
개선 제안: 현재 구현이 적절해 보입니다.