如果我们列出10以下的所有自然数是3或5的倍数,我们得到3,5,6和9.这些倍数的总和是23。 完成解决方案,使其返回传入数字下方所有3或5的倍数之和。

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in.

代码:

import java.util.stream.IntStream;

public class SolutionBei {
    //方法一
    public static int solution(int number) {
        int sum=0;
        for(int i=1;i<number;i++){
            int a=i%3;
            int b=i%5;
            if(a==0||b==0){
                sum=sum+i;
            }
        }
        return sum;
    }
    //方法二
    public static int solution2(int number) {
        return IntStream.range(0, number)
                .filter(n -> (n % 3 == 0) || (n % 5 == 0))
                .sum();
    }
    public static void main(String[] args) {
        System.out.println(solution(10));
        System.out.println(solution2(10));
    }
}

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转载自blog.csdn.net/weixin_43969830/article/details/89282006
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