年龄计算总是差一

Age computation always off by one

任务是创建函数。

该函数有两个参数:

计算多少年前父亲的年龄是儿子的两倍(或多少年后他将是儿子的两倍)。

public static int TwiceAsOld(int dadYears, int sonYears){

    int dadYearsTemp = dadYears;
    int years = 0;
    int yearsAgo = 0;

    for (int i = 0; i <= dadYears; i++ ){
        if (dadYearsTemp / 2 == sonYears) {
            years = dadYearsTemp;
            yearsAgo = dadYears - years;
            System.out.println(yearsAgo);
            return yearsAgo;
        }
        else if (sonYears * 2 > dadYears) {
            years = (sonYears * 2) - dadYears;
            System.out.println(years);
            return years;
        }
        dadYearsTemp = dadYearsTemp -1;
    }

    return 42; // The meaning of life

}

例如,输入 (30, 7) 我希望我的函数为 return 16,因为 16 年前父亲是 14,这意味着他现在是他儿子的两倍大(7).但是我的函数 returns 15.

我想这不是一个大错误,但老实说我无法找出它不起作用的原因,所以我希望得到一些帮助。

让爸爸现在的年龄=f

儿子现在的年龄=s

让x年前,父亲是儿子的两倍。

然后 2(s-x) = (f-x) => x = 2*s - f

注意:如果x为负数,那么x年后,父亲的年龄是儿子的两倍(测试输入[25,5])

public static void main(String[] args) {
    int x = twiceAsOld(25, 5);
    System.out.println(x);
  }

  public static int twiceAsOld(int dadYears, int sonYears) {
    return 2*sonYears - dadYears;

  }

下面是示例代码,用于计算多少年前父亲的年龄是儿子的两倍,带有示例数据。

 public static void main (String args[]) {
    int sonAge = 10;
    int fatherAge = 24;
    int yearsAgo = 0;
    while(sonAge*2 != fatherAge && fatherAge>0 ) {
        yearsAgo++;
        fatherAge = fatherAge-1;
    }
    System.out.println(yearsAgo);
}