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[grapefruit13] WEEK 1 Solutions #2359
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,16 @@ | ||
| /** | ||
| * @description nums 배열에서 중복 숫자 확인 | ||
| * @param nums - 숫자 배열 | ||
| * @returns boolean - 중복 숫자 여부 | ||
| */ | ||
| const containsDuplicate = (nums: number[]) => { | ||
| const hasSeen = new Set<number>(); | ||
|
|
||
| for (const num of nums) { | ||
| if (hasSeen.has(num)) { | ||
| return true; | ||
| } | ||
| hasSeen.add(num); | ||
| } | ||
| return false; | ||
| }; |
|
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. 해시맵을 이용한 깔끔한 풀이법이라고 생각합니다! 👍 |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,18 @@ | ||
| /** | ||
| * @description nums 배열에서 두 수를 더해 target을 만족하는 인덱스를 반환 | ||
| * @param {number[]} nums - 숫자 배열 | ||
| * @param {number} target - 목표 숫자 | ||
| * @return {number[]} - 두 수의 인덱스 | ||
| */ | ||
| var twoSum = function (nums, target) { | ||
| const map = new Map(); | ||
|
|
||
| for (let i = 0; i < nums.length; i++) { | ||
| const current = nums[i]; | ||
| const needed = target - current; | ||
| if (map.has(needed)) { | ||
| return [map.get(needed), i]; | ||
| } | ||
| map.set(current, i); | ||
| } | ||
| }; |
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Set을 이용해서 깔끔하게 풀이하셨네요. 시간 / 공간 복잡도에 가장 최적화된 방법인 것 같습니다. 👍