feat: 압축-카카오 문제 reduce 활용 풀이 추가 #102
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codeisneverodd
approved these changes
Sep 21, 2022
level-2/[3차]-압축&17684&.js
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| const dict = [ | ||
| "A", | ||
| "B", | ||
| "C", | ||
| "D", | ||
| "E", | ||
| "F", | ||
| "G", | ||
| "H", | ||
| "I", | ||
| "J", | ||
| "K", | ||
| "L", | ||
| "M", | ||
| "N", | ||
| "O", | ||
| "P", | ||
| "Q", | ||
| "R", | ||
| "S", | ||
| "T", | ||
| "U", | ||
| "V", | ||
| "W", | ||
| "X", | ||
| "Y", | ||
| "Z", | ||
| ]; |
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직접 적어주는 방식은 알파벳이 누락되거나, 대소문자등의 문제가 생길 수 있습니다!
다음과 Array.from과 아스키코드를 사용한 방식은 어떨까요?
const dict = Array.from({length: 26},(_, i) => String.fromCharCode(65 + i))
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어우 너무 좋은 방법같습니다 :) 👍
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| const lastWordAndCompression = msg.split("").reduce((acc, cur) => { | ||
| const nextWord = acc + cur; | ||
| const nextWordIdx = dict.indexOf(nextWord); | ||
| const prevWordIdx = dict.indexOf(acc); | ||
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| if (nextWordIdx !== -1) return acc + cur; | ||
| dict.push(nextWord); | ||
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| if (prevWordIdx !== -1) result.push(prevWordIdx + 1); | ||
| return cur; | ||
| }, ""); |
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와우 너무 멋진 풀이네요!!! reduce 메서드를 이번 풀이에 멋지게 사용하신 것 같습니다 배워갑니다!! 😄
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기존 풀이에 추가한 풀이
레벨 2, [3차]압축 풀이 추가: 9c743c6