题解 | #质数因子#

这个问题大部分的解法都是这样，利用循环穷举。

Scanner scanner = new Scanner(System.in);

long num = scanner.nextLong();
while(num != 1)
{
            for (int i = 2; i <= num; ++i) {
                if (num % i == 0) {
                    System.out.print(i + " ");
                    num /= i;
                    break;
                }
            }
}

也不能说错，但是这个方法遇到大数就过不了。
所以要怎么优化呢？
1.降低循环次数，利用Math.sqrt

long k = (long) Math.sqrt(num);

2.利用双重循环，假如输入的数值能被2整除，那就一直整除下去

for (long i = 2; i <= k; ++i) {
            while (num % i == 0) {
                System.out.print(i + " ");
                num /= i;
            }
}

3.最后再判断除完的数有没有大于1

System.out.println(num == 1 ? "": num+" ");

那么最终的解法是

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        long num = scanner.nextLong();
        long k = (long) Math.sqrt(num);

        for (long i = 2; i <= k; ++i) {
            while (num % i == 0) {
                System.out.print(i + " ");
                num /= i;
            }
        }
        System.out.println(num == 1 ? "": num+" ");
    }
}