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[moonhyeok] Week 4 #840

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merged 2 commits into from
Jan 5, 2025
Merged

[moonhyeok] Week 4 #840

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59e8a48
feat: Upload merge-two-sorted-lists (typescript)
mike2ox Jan 4, 2025
be71d9f
feat: Upload missing-number(typescript)
mike2ox Jan 4, 2025
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43 changes: 43 additions & 0 deletions merge-two-sorted-lists/mike2ox.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,43 @@
/**
* source: https://leetcode.com/problems/merge-two-sorted-lists/
* 풀이방법: 두 리스트를 비교하면서 작은 값을 결과 리스트에 추가
*
* 시간복잡도: O(n + m) (n: list1의 길이, m: list2의 길이)
* 공간복잡도: O(1) (상수 공간만 사용)
*
*/

class ListNode {
val: number;
next: ListNode | null;
constructor(val?: number, next?: ListNode | null) {
this.val = val === undefined ? 0 : val;
this.next = next === undefined ? null : next;
}
}

function mergeTwoLists(
list1: ListNode | null,
list2: ListNode | null
): ListNode | null {
const result = new ListNode();
let current = result;
while (list1 !== null && list2 !== null) {
if (list1.val <= list2.val) {
current.next = list1;
list1 = list1.next;
current = current.next;
} else {
current.next = list2;
list2 = list2.next;
current = current.next;
}
}
if (list1 !== null) {
current.next = list1;
}
if (list2 !== null) {
current.next = list2;
}
return result.next; // 처음에 추가한 더미 노드 제외
}
15 changes: 15 additions & 0 deletions missing-number/mike2ox.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
/**
* source: https://leetcode.com/problems/missing-number/
* 풀이방법: 0부터 n까지의 합에서 주어진 배열의 합을 빼면 빠진 숫자를 구할 수 있음
*
* 시간복잡도: O(n) (n: nums의 길이)
* 공간복잡도: O(1) (상수 공간만 사용)
*/

function missingNumber(nums: number[]): number {
const n = nums.length;
let expectedSum = (n * (n + 1)) / 2; // 0부터 n까지의 합 공식
let realSum = nums.reduce((sum, cur) => sum + cur, 0);

return expectedSum - realSum;
}
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