[Coding test] 0328_no.3 (java)
2020-03-28
Coding test / 0328_no.3
문제설명
유출 x
풀이
package T0328;
public class t3 {
public static int solution(int A[]) {
String s1 = "", s2 = "", s3= "", s4 = "";
for(int i=0; i<A.length-1; i++) {
if(i%2 == 0) {
s1 += "0";
s2 += "1";
}
else {
s1 += "1";
s2 += "0";
}
}
for(int i=0; i<A.length-2; i++) {
if(i%2 == 0) {
s3 += "0";
s4 += "1";
}
else {
s3 += "1";
s4 += "0";
}
}
//지우지 않고도 조건을 만족하는지
String a = ""; int answer = -1; int cnt = 0;
for(int i=1; i<A.length; i++) {
if(A[i] > A[i-1])
a += "0";
else if(A[i] < A[i-1])
a += "1";
else
a += "-";
}
if(a.charAt(0) == '0' && a.equals(s1))
return 0;
else if(a.charAt(0) == '1' && a.equals(s2))
return 0;
//아닌 경우에
for(int i=0; i<A.length; i++) {
int arr[] = new int[A.length-1];
int k = 0;
for(int j=0; j<A.length; j++) {
if(i == j)
continue;
arr[k++] = A[j];
}
String s = "";
for(int j=1; j<A.length-1; j++) {
if(arr[j] > arr[j-1])
s += "0";
else if(arr[j] < arr[j-1])
s += "1";
else
s += "-";
}
if(s.charAt(0) == '0' && s.equals(s3))
cnt++;
else if(s.charAt(0) == '1' && s.equals(s4))
cnt++;
}
if(cnt != 0)
answer = cnt;
return answer;
}
public static void main(String[] args) {
int a[] = {3,4,5,3,7};
int b[] = {1,2,3,4};
int c[] = {1,3,1,2};
int d[] = {1,1,1,1,1};
System.out.println(solution(d));
}
}