웹개발자
태그 목록
- 전체보기(69)
- js(9)
- split()(3)
- toUpperCase(2)
- reduce(2)
- (Two Pointers Algorithm)(2)
- for ~ of(2)
- substring(2)
- replace(2)
- 알고리즘(1)
- JOIN(1)
- queue.push(queue.shift())(1)
- charCodeAt(1)
- Arrays.stream(arr).max().getAsInt();(1)
- filter(1)
- two pointer(1)
- foreach(1)
- Math함수(1)
- toLowerCase(1)
- 문자열 자르기(1)
- 정규식(1)
- reverse(1)
- map.has()(1)
- new Map()(1)
- substr(1)
- isNaN(1)
- toString(1)
- slice(1)
- Character.isAlphabetic(1)
- Arrays.stream(arr).sum();(1)
- Array.from(1)
- for 반복문(1)
- indexOf(1)
- Array.from({length: n}, (v, i) => i+1)(1)
- for문(1)
- isPrime(1)
- [arr[i], arr[idx]] = [arr[idx], arr[i]];(1)
- Sliding Window(1)
전체보기 (69)js(9)split()(3)toUpperCase(2)reduce(2)(Two Pointers Algorithm)(2)for ~ of(2)substring(2)replace(2)알고리즘(1)JOIN(1)queue.push(queue.shift())(1)charCodeAt(1)Arrays.stream(arr).max().getAsInt();(1)filter(1)two pointer(1)foreach(1)Math함수(1)toLowerCase(1)문자열 자르기(1)정규식(1)reverse(1)map.has()(1)new Map()(1)substr(1)isNaN(1)toString(1)slice(1)Character.isAlphabetic(1)Arrays.stream(arr).sum();(1)Array.from(1)for 반복문(1)indexOf(1)Array.from({length: n}, (v, i) => i+1)(1)for문(1)isPrime(1)[arr[i], arr[idx]] = [arr[idx], arr[i]];(1)Sliding Window(1)
중복 단어 제거
나의 풀이 function solution(arr) { let answer; answer = arr.filter((v,i)=>{ if(arr.indexOf(v) === i) return v; // 해당되는 원소(v) 반환 }) return answer; } let arr = ['good', 'time', 'goo...
2021년 6월 4일
·
0개의 댓글