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dohyeon2
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Mar 12, 2026
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| let mut arr: [i32; 2] = [1, 2]; | ||
| let mut current = 0; | ||
| for n in 3..=n { |
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변수명이 반복되어 오해의 소지가 있는 것 같습니다!
사용되지 않으니 언더바로 처리하는건 어떨까요?
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그러게요. 사용안하고 있었네요. 감사합니다.
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| target := nums[k] | ||
| for left < right { |
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반본묵 문법이 신기하네요
옛~날에 배운걸 떠올려보면 반복문이 for문으로 통일되어 있던것 같은데 맞나요?
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네 맞습니다. 이 경우는 while 문으로 보시면 될 것 같아요.
| for left < right { | ||
| sum := nums[left] + nums[right] + target | ||
| if sum == 0 { | ||
| if left > k { |
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중복제거를 key 통일 방식으로 해결하셨군요
근데 이렇게하면 중복되는 케이스들도 연산이 필수적이되어서
중복확인을 먼저수행하는 케이스보다 효율이 조금 떨어질 것 같은데
어떻게 생각하시나요?
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sum 계산하기 전에 key 중복 확인을 먼저하는게 효율이 좋을거 같다는 말씀이시죠?
hyejj19
approved these changes
Mar 13, 2026
Comment on lines
+3
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| func productExceptSelf(nums []int) []int { | ||
| result := make([]int, 0) | ||
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| left := make([]int, len(nums)) | ||
| left[0] = 1 | ||
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| right := make([]int, len(nums)) | ||
| right[len(nums)-1] = 1 | ||
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| for i := 1; i < len(nums); i++ { | ||
| left[i] = nums[i-1] * left[i-1] | ||
| } | ||
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| for i := len(nums) - 2; i >= 0; i-- { | ||
| right[i] = nums[i+1] * right[i+1] | ||
| } | ||
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| for i := 0; i < len(nums); i++ { | ||
| result = append(result, left[i]*right[i]) | ||
| } | ||
| return result | ||
| } |
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오! left와 right 배열을 각각 만들어서 계산하셨군요.
left 결과가 담긴 배열에 오른쪽 누적곱을 바로 곱해주면 배열 하나로도 풀 수 있을 것 같아요! (문법이 엄청 신기하네요 ㅎㅎ)
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