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Mar 6, 2026
| # Time Complexity: O(n) (Set) | ||
| class Solution: | ||
| def containsDuplicate(self, nums: List[int]) -> bool: | ||
| return len(nums) != len(set(nums)) |
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저랑 동일하게 푸셨네요!
저도 좀 더 찾아보았는데요 저희가 푼 방식은 조기 종료가 되지 않아서 아래 처럼 조기 종료하면 좀더빠르게 순회를 종료할꺼같아요!
조기 종료가 안 됨 — Set 방식은 항상 전체 배열을 순회합니다. 중복이 앞쪽에 있어도 끝까지 돌아야 하죠.
var containsDuplicate = function (nums) {
const seen = new Set();
for (const n of nums) {
if (seen.has(n)) return true;
seen.add(n);
}
return false;
};
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전혀 생각하지 못했던 부분이네요, 감사합니다.
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