n的阶乘

public class jiecheng {
@Test
public void ss() {
System.out.println(aa(4));
}
public int aa(int n ){
if(n==0){
return 1;
}else{
long num =n*aa(n-1);
return (int) num;
}
}

}

转载于:https://www.cnblogs.com/itcx1213/p/10963783.html