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답변 형식
common 배열을 받아 원소들의 관계를 등차수열 또는 등비수열로 판단한다.
등비 또는 등차의 크기를 구하여 마지막 원소 다음에 나올 숫자를 찾는다.
풀이 조건
최소 3개 이상의 원소가 나오기 때문에
– 1항과 2항
– 2항과 3항의 관계를
확인하면 됩니다.
배열의 크기가 999까지 가능하지만
종항의 다음 항이 int 범위 내에 있기 때문에 문제를 풀 때 형 크기에 대한 부담은 없습니다.
JAVA 정답
class Solution {
public int solution(int[] common) {
int answer = 0;
int size = common.length;
int lastnum = size-1;
if(common[1]-common[0] == common[2]-common[1])
answer = common[lastnum]+(common[1] – common[0]);
else
answer = common[lastnum]*(common[1]/common[0]);
return answer;
}
}
JAVA 정답 해설
class Solution {
public int solution(int[] common) {
int answer = 0;
int size = common.length;
/// 배열의 크기를 구한다.
int lastnum = size-1;
/// 배열의 크기가 10이라면 마지막 항은 9항이다.(배열은 0항 시작)
if(common[1]-common[0] == common[2]-common[1])
/// 1항과 0항의 뺄셈 값 차이가 2항과 1항의 뺄셈 값 차이와 같다면 등차 수열로 판단한다.
answer = common[lastnum]+(common[1] – common[0]);
/// 등차수열이라면 마지막 항에 등차를 더해서 제출한다.
else
/// 배열이 등차수열이 아니라면 등비수열이다.
answer = common[lastnum]*(common[1]/common[0]);
/// 등비수열이라면 마지막 항에 등비를 곱해서 제출한다.
return answer;
}
}
JAVA 간결한 정답
class Solution {
public int solution(int[] common) {
if(common[1]-common[0] == common[2]-common[1])
return common[common.length–1]+(common[1] – common[0]);
else return common[common.length–1]*(common[1]/common[0]);
}
}
JAVA 삼항연산자
class Solution {
public int solution(int[] common) {
return common[1]-common[0] == common[2]-common[1] ?
common[common.length–1]+(common[1] – common[0]) :
common[common.length–1]*(common[1]/common[0]);
}
}
Python3 정답
def solution(common):
if common[1] – common[0] == common[2] – common[1] :
return common[-1]+(common[1] – common[0])
else :
return common[-1]*(common[1]/common[0])
Python3 정답 해설
def solution(common):
if common[1] – common[0] == common[2] – common[1] :
# 1항과 0항의 뺄셈 값 차이가 2항과 1항의 뺄셈 값 차이와 같다면 등차 수열로 판단한다.
return common[-1]+(common[1] – common[0])
# 등차수열이라면 마지막 항에 등차를 더해서 제출한다.
else :
# 배열이 등차수열이 아니라면 등비수열이다.
return common[-1]*(common[1]/common[0])
# 등비수열이라면 마지막 항에 등비를 곱해서 제출한다.
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