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feat: 문제풀이 추가
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JEONGHWANMIN committed Oct 3, 2024

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commit 07e09296b16bb393737e8abe152aa8ec1f026da7
37 changes: 37 additions & 0 deletions merge-two-sorted-lists/hwanmini.js
Original file line number Diff line number Diff line change
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// 시간복잡도: O(m + n)
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clone graph 문제에서 해주신 것처럼 m, n을 무엇으로 정의했는지 적어주면 더 좋을 것 같아요.

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좋습니다 ~

// 공간복잡도: O(m + n)

/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} list1
* @param {ListNode} list2
* @return {ListNode}
*/
var mergeTwoLists = function(list1, list2) {
let res = new ListNode()
let resCopy = res

while (list1 && list2) {
if (list1.val < list2.val) {
res.next = new ListNode(list1.val);
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지금은 순회를 할때마다 새로운 노드를 만들어서 할당하고 있어요.
이걸 개선할 수 있는 방법이 있을까요~? 🤔

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아 기존 리스트 노드를 재활용하면 되겠네요..!
감사합니다 : )

list1 = list1.next;
} else {
res.next = new ListNode(list2.val);
list2 = list2.next;
}

res = res.next
}

if (list1) res.next = list1;
if (list2) res.next = list2

return resCopy.next
};