ContainsDuplicate #3
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Two Sum
정석 포문으로 풀어보았습니다. |
lukasjhan
requested changes
Feb 17, 2024
TwoSum.js
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Comment on lines
8
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18
| for (let i = 0; i < nums.length; i++) { | ||
| const x = nums[i]; | ||
| for (let j = i + 1; j < nums.length; j++) { | ||
| const v = nums[j]; | ||
| const sum = x + v; | ||
|
|
||
| if (sum === target) { | ||
| answer = [i, j]; | ||
| break; | ||
| } | ||
| } |
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이중 for문으로 풀면 안됩니다. HashMap 개념을 사용해야해요.
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이중 for문으로 풀면 시간복잡도가 O(n^2)이 됩니다.
|
target(더한 목표 값)에서 현재 반복 돌리고 있는 숫자를 뺀 값을 변수에 담습니다. |
lukasjhan
approved these changes
Feb 17, 2024
|
@yunseorim1116 최고에요 ㅋㅋㅋ |
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ContainsDuplicate
객체를 검사하면서 2이상의 숫자가 있으면 바로 true 를 리턴하게 짜도 됐지만
12 : 객체 변환, 14 : 검사
역할을 한줄로 각각 분리하고 싶어서 나누어보았습니다.