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[jj7779607] Week4 #827
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[jj7779607] Week4 #827
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,29 @@ | ||
/** | ||
* 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) { | ||
// 리스트가 비었을 때 다른 리스트 반환 | ||
if (list1 === null) return list2; | ||
if (list2 === null) return list1; | ||
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// 작은 값 가진 노드 선택하고 재귀호출 | ||
if (list1.val <= list2.val) { | ||
list1.next = mergeTwoLists(list1.next, list2); | ||
return list1; | ||
} else { | ||
list2.next = mergeTwoLists(list1, list2.next); | ||
return list2; | ||
} | ||
}; | ||
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// 시간 복잡도: O(n1+n2) | ||
// 공간 복잡도: O(1) |
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. 정렬 후 인덱스 비교하는 건 좋은 아이디어인 것 같습니다. 공간복잡도는 O(1)으로 유지하고 시간복잡도는 O(n) 으로 줄이기 위해, 수학적 합 공식을 활용해봐도 좋을 것 같아요! 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. 시간복잡도를 줄이기 위해 수학적 합 공식 한번 활용해봐야겠네요! 감사합니다! |
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/** | ||
* @param {number[]} nums | ||
* @return {number} | ||
*/ | ||
var missingNumber = function (nums) { | ||
// 1. nums 정렬 | ||
nums.sort((a, b) => a - b); | ||
// 2. for문 돌며 빠진 숫자 찾기 | ||
for (let i = 0; i <= nums.length; i++) { | ||
if (nums[i] !== i) { | ||
return i; | ||
} | ||
} | ||
}; | ||
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// 시간 복잡도: O(nlogn) | ||
// 공간 복잡도: O(1) |
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리뷰가 너무 늦어 죄송합니다 ^ㅠ 재귀로 링크드 리스트를 병합하는 방법을 잘 배우고 갑니다. 이번 주도 수고하셨습니다!