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feat: missing-number 알고리즘 리팩토링
- set 자료구조를 활용 - for문을 모두 순회한경우 리스트의 길이를 리턴
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missing-number/jinah92.py

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@@ -1,12 +1,10 @@
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# O(n) times, O(n) spaces
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class Solution:
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def missingNumber(self, nums: List[int]) -> int:
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nums_keys = dict.fromkeys(nums,0)
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last = 0
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nums_keys = set(nums)
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for i in range(len(nums)):
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if not i in nums_keys:
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return i
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last += 1
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return last
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return len(nums)

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