[Tessa1217] Week 11 Solutions #1569
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uraflower
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Jun 13, 2025
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| /** | ||
| * 0 ~ n까지 고유한 숫자들로 이루어진 배열 nums가 주어질 때 해당 범위 내에서 없어진 숫자를 찾으세요. | ||
| */ | ||
| class Solution { | ||
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| // 시간복잡도: O(n), 공간복잡도: O(n) | ||
| public int missingNumber(int[] nums) { | ||
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| boolean[] visitNum = new boolean[nums.length + 1]; | ||
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| for (int i = 0; i < nums.length; i++) { | ||
| visitNum[nums[i]] = true; | ||
| } | ||
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| for (int i = 0; i < visitNum.length; i++) { | ||
| if (!visitNum[i]) { | ||
| return i; | ||
| } | ||
| } | ||
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| return 0; | ||
| } | ||
| } | ||
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0부터 n까지 모든 숫자가 포함된다는 걸 이용해, 0부터 n까지 모두 더한 값과 주어진 배열의 모든 원소를 더한 값의 차를 반환하는 방법도 있습니다. 이렇게 하면 배열에 저장하지 않아도 돼서 공간복잡도를 O(1)로 개선할 수 있어서 참고하시면 좋을 것 같습니다!
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