[jinho] Week 1 #632
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[jinho] Week 1 #632
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obzva
reviewed
Dec 5, 2024
top-k-frequent-elements/neverlish.go
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Comment on lines
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| freq_by_counts := make(map[int][]int) | ||
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| for num, count := range freq { | ||
| if _, ok := freq_by_counts[count]; !ok { | ||
| freq_by_counts[count] = []int{} | ||
| } | ||
| freq_by_counts[count] = append(freq_by_counts[count], num) | ||
| } | ||
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| result := []int{} | ||
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| for count := len(nums); count > 0; count-- { | ||
| if nums, ok := freq_by_counts[count]; ok { | ||
| if len(result) >= k { | ||
| break | ||
| } | ||
| result = append(result, nums...) | ||
| } | ||
| } |
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문제의 조건을 잘 이용한 풀이라고 생각합니다 :)
It is guaranteed that the answer is unique.
obzva
reviewed
Dec 5, 2024
top-k-frequent-elements/neverlish.go
Outdated
| freq_by_counts[count] = append(freq_by_counts[count], num) | ||
| } | ||
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| result := []int{} |
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result slice를 선언할 때 make를 이용해서 capacity나 length를 설정해주는 것이 더 효율적일 것 같습니다
k가 커지면 re-allocation에 소모되는 연산량이 많을 수도 있을 것 같아요
Suggested change
| result := []int{} | |
| result := make([]int, k) // Or, result := make([]int, 0, k) |
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487db4c 요렇게 작성해보았습니다
obzva
reviewed
Dec 5, 2024
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피드백 감사합니다! |
obzva
approved these changes
Dec 12, 2024
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풀이 잘 보았습니다..! 제가 go언어는 처음봐서 좀 더 유심히 보고 분석하느라 재밋었습니다! 아직은 코드가 익숙하지않아서 좋은 공부시간이였던 것 같습니다 수고하셨어요! |
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