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Merge pull request #41 from prove-ability/main

[정기적 풀이 추가] 2022年04月26日

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`@@ -24,3 +24,13 @@ function solution(s) {`
`24`	`24`	`// 문자열에서 특정 문자의 개수를 구하려면 split을 사용하면 된다.`
`25`	`25`	`// Ex. "ababb".split("a") 의 결과는 ["", "b", "bb"]`
`26`	`26`	`// => 즉, "a"의 갯수는 3에서 1을 뺀 2`
	`27`	`+`
	`28`	`+// 정답 4 - prove-ability`
	`29`	`+function solution(s){`
	`30`	`+ // 배열로 변환`
	`31`	`+ s = s.split("");`
	`32`	`+ // filter 를 사용해 갯수 추출`
	`33`	`+ const pCount = s.filter((v) => v === "p" \|\| v === "P").length;`
	`34`	`+ const yCount = s.filter((v) => v === "y" \|\| v === "Y").length;`
	`35`	`+ return pCount === yCount;`
	`36`	`+}`

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`@@ -56,4 +56,25 @@ function solution(n) {`
`56`	`56`	`if (isPrime(i)) answer++;`
`57`	`57`	`}`
`58`	`58`	`return answer;`
	`59`	`+}`
	`60`	`+`
	`61`	`+//정답 3 - prove-ability`
	`62`	`+// 소수 판별 로직`
	`63`	`+function isPrime(n) {`
	`64`	`+ // n 제곱근 후 올림`
	`65`	`+ for (let i = 2, len = Math.ceil(Math.sqrt(n)); i <= len; i++) {`
	`66`	`+ if (n % i === 0) return false;`
	`67`	`+ }`
	`68`	`+ return true;`
	`69`	`+}`
	`70`	`+`
	`71`	`+function solution(n) {`
	`72`	`+ let count = 0;`
	`73`	`+ // 1부터 n까지 반복적으로 접근 - i`
	`74`	`+ for(let i = 1; i <= n; i++) {`
	`75`	`+ // i 가 소수인지 확인 후 count++`
	`76`	`+ if(isPrime(i)) count++;`
	`77`	`+ }`
	`78`	`+`
	`79`	`+ return count;`
`59`	`80`	`}`

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`@@ -40,4 +40,21 @@ function solution(left, right) {`
`40`	`40`	`return answer;`
`41`	`41`	`}`
`42`	`42`
`43`		`-//`
	`43`	`+// 정답 4 - prove-bility`
	`44`	`+function getDivisorCount(i) {`
	`45`	`+ let count = 0;`
	`46`	`+ for(let j = 1; j <= i; j++) {`
	`47`	`+ if(i % j === 0) count++;`
	`48`	`+ }`
	`49`	`+ return count;`
	`50`	`+}`
	`51`	`+`
	`52`	`+function solution(left, right) {`
	`53`	`+ let answer = 0;`
	`54`	`+ for(let i = left; i <= right; i++) {`
	`55`	`+ let count = getDivisorCount(i);`
	`56`	`+ if(count % 2 === 0) answer += i;`
	`57`	`+ else answer -= i;`
	`58`	`+ }`
	`59`	`+ return answer;`
	`60`	`+}`

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`@@ -42,3 +42,13 @@ function solution(absolutes, signs) {`
`42`	`42`
`43`	`43`	`return answer;`
`44`	`44`	`}`
	`45`	`+`
	`46`	`+// 정답 5 - prove-ability`
	`47`	`+function solution(absolutes, signs) {`
	`48`	`+ let answer = 0;`
	`49`	`+ absolutes.forEach((absolute, i) => {`
	`50`	`+ if(!signs[i]) absolute *= -1;`
	`51`	`+ answer += absolute;`
	`52`	`+ })`
	`53`	`+ return answer;`
	`54`	`+}`

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`@@ -12,3 +12,13 @@ function solution(n) {`
`12`	`12`	`// x가 정수이면 x+1의 제곱 반환, x가 정수가 아니면 -1 반환`
`13`	`13`	`return Number.isInteger(x) ? Math.pow(x + 1, 2) : -1;`
`14`	`14`	`}`
	`15`	`+`
	`16`	`+//정답 3 - prove-ability`
	`17`	`+function solution(n) {`
	`18`	`+ // n의 제곱근을 x 초기화`
	`19`	`+ const x = Math.sqrt(n);`
	`20`	`+ // 양의 정수라면 x + 1 제곱 반환`
	`21`	`+ if(Number.isInteger(x)) return Math.pow(x + 1, 2)`
	`22`	`+ // 아니라면 -1 반환`
	`23`	`+ return -1;`
	`24`	`+}`

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