[Tessa1217] WEEK 01 solutions #1129
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grapefruitgreentealoe
approved these changes
Apr 4, 2025
| Set<Integer> distincts = new HashSet<>(); | ||
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| for (int i = 0; i < nums.length; i++) { | ||
| distincts.add(nums[i]); |
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중복을 찾는 것이 목적이므로, 전체 배열을 다 순회하지 않고 중복 발견 즉시 리턴하는 방식은 어떨까요?
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`
public boolean containsDuplicate(int[] nums) {
Set<Integer> distincts = new HashSet<>();
for (int i = 0; i < nums.length; i++) {
boolean added = distincts.add(nums[i]);
if (!added) {
return true;
}
}
return false;
}
`
리뷰해주신대로 개선했더니 확실히 런타임이 줄었네요. 좋은 리뷰 감사합니다!
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| for (int i = 0; i < nums.length; i++) { | ||
| if (numMap.containsKey(target - nums[i])) { | ||
| answer[0] = numMap.get(target - nums[i]); |
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target - nums[i]에 대한 변수를 생성해서 연산한 값을 담아주면 가독성과 연산 최소화가 가능해보입니다!
| if (numMap.containsKey(target - nums[i])) { | ||
| answer[0] = numMap.get(target - nums[i]); | ||
| answer[1] = i; | ||
| break; |
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배열은 조건 만족 시점에 생성하고 break없이 바로 리턴해도 될것같습니다.
| return 0; | ||
| } | ||
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| Arrays.sort(nums); |
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문제 제약조건이 시간복잡도 O(n) 까지 허용인데 혹시 통과되셨나요?
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넵 시간복잡도가 정렬 때문에 O(n log n)이 될 것 같기는 한데 일단 문제 풀이 자체는 통과했습니다. 그래도 O(n)으로 시간 복잡도를 줄일 수 있는 풀이를 한번 찾아보는 게 좋겠네요. 좋은 리뷰 감사합니다! 1주차 고생하셨습니다!
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