我正在努力获得一个数字的所有主要因素。 for
循环应该一直有效,直到找到匹配并且它应该中断并跳转到下一个if
语句,该语句检查数字是否不等于零。
public class Factor {
public static ArrayList <Integer> HoldNum = new ArrayList();
public static void main(String[]args){
Factor object = new Factor();
object.Factor(104);
System.out.println(HoldNum.get(0));
}
public static int Factor(int number){
int new_numb = 0;
int n=0;
for( n = 1; n < 9; n++) {
if (number % n == 0) {
HoldNum.add(n);
new_numb = number/n;
break;
}
}
System.out.println(new_numb);
if(new_numb < 0) {
HoldNum.add(new_numb);
return 1;
} else {
return Factor(new_numb);
}
}
}
答案 0 :(得分:1)
至少有三个错误:
答案 1 :(得分:0)
您要求提供递归解决方案。你走了:
public class Example {
public static void main(String[] args) {
System.out.println(factors(104));
}
public static List<Integer> factors(int number) {
return factors(number, new ArrayList<Integer>());
}
private static List<Integer> factors(int number, List<Integer> primes) {
for (int prim = 2; prim <= number; prim++) {
if (number % prim == 0) {
primes.add(prim);
return factors(number / prim, primes);
}
}
return primes;
}
}
代码不是防弹的,它只是一个快速而肮脏的例子。
答案 2 :(得分:0)
Java实现......
public class PrimeFactor {
public int divisor=2;
void printPrimeFactors(int num)
{
if(num == 1)
return;
if(num%divisor!=0)
{
while(num%divisor!=0)
++divisor;
}
if(num%divisor==0){
System.out.println(divisor);
printPrimeFactors(num/divisor);
}
}
public static void main(String[] args)
{
PrimeFactor obj = new PrimeFactor();
obj.printPrimeFactors(90);
}
}