The question is very short and the solution is also very convenient; this is also a question that you will not know what to do if you just start to study the question, and you need to be patient and carefully look at the information. Take a look at the conditions, the array is sorted, and the solution can be easily obtained.
I think it's just an output length, so I simply counted the lengths of different numbers in the array without performing any delete operations, and it was done on PTA.
From a respectful point of view and an honest attitude, I symbolically deleted the answer after outputting it, denoted by -1.
package delDulplicate;
import java.util.Scanner;
//输入数据到数组
//输入的数据本来就是排好序的
//如果当前位和前一位一样 后面提上来
public class Main {
static final int MAXN= 110;
static int n;
static int[] array= new int[MAXN];
public static void main(String[] args) {
Scanner console= new Scanner(System.in);
n= console.nextInt();
for(int i=0; i<n; i++) {
array[i]= console.nextInt();
}
int count=1;
for(int i=0; i<n-1; i++) {
if(array[i]!=array[i+1])
count++;
}
System.out.println(count);
//执行重复删除操作
for(int i=n-1; i>=1; i--) {
if(array[i]==array[i-1])
move(i);
}
for(int i=count; i<n; i++) {
array[i]= -1;
}
// for(int i=0; i<n; i++) {
// System.out.println(array[i]);
// }
}
//将i之后的都向前移动一位
public static void move(int i) {
for(int j=i; j<n-1; j++) {
array[j]= array[j+1];
}
}
}