找到两个日期之间的差异的方法

Question

我的方法是有效的，因为我只想知道几年之间的差异，我的问题是我这样做效率低下。我有一种感觉，写出这种方法有一种更加简单和美观的方式。

编辑：我必须写我自己的方法。我也更喜欢不要走出自己的路，并使用先进的东西，在一年级程序员的范围内。

public int differenceInYears(MyDate comparedDate) { 
     int difference = 0; 
     if (this.year > comparedDate.year) { 
      if (this.month > comparedDate.month) { 
       difference = this.year - comparedDate.year; 
      } 
      else if (this.month == comparedDate.month) { 
       if (this.day >= comparedDate.day) { 
        difference = this.year - comparedDate.year; 
       } 
       else { 
        difference = this.year - comparedDate.year - 1; 
       } 
      } 
      else { 
       difference = this.year - comparedDate.year - 1; 
      } 
     } 
     if (comparedDate.year > this.year) { 
      if (comparedDate.month > this.month) { 
       difference = comparedDate.year - this.year; 
      } 
      else if (comparedDate.month == this.month) { 
       if (comparedDate.day >= this.day) { 
        difference = comparedDate.year - this.year; 
       } 
       else { 
        difference = comparedDate.year - this.year - 1; 
       } 
      } 
      else { 
       difference = comparedDate.year - this.year - 1; 
      } 
     } 
     return difference; 
    } 

我将添加MyDate类下面的背景：

public class MyDate { 

    private int day; 
    private int month; 
    private int year; 

    public MyDate(int day, int montd, int year) { 
     this.day = day; 
     this.month = montd; 
     this.year = year; 
    } 

    public String toString() { 
     return this.day + "." + this.month + "." + this.year; 
    } 

    public boolean earlier(MyDate compared) { 
     if (this.year < compared.year) { 
      return true; 
     } 

     if (this.year == compared.year && this.month < compared.month) { 
      return true; 
     } 

     if (this.year == compared.year && this.month == compared.month 
       && this.day < compared.day) { 
      return true; 
     } 

     return false; 
    } 

Answer 1

我会假设你的方法正常工作。如何改进？您可以做的第一件事是消除所有重复的this.year - comparedDate.year和comparedDate.year - this.year计算。无论做什么，你都要做，所以让我们把它们放在它们各自的if块的顶部。

public int differenceInYears(MyDate comparedDate) { 
    int difference; 

    if (this.year > comparedDate.year) { 
     difference = this.year - comparedDate.year; 

     if (this.month > comparedDate.month) { 
     } 
     else if (this.month == comparedDate.month) { 
      if (this.day >= comparedDate.day) { 
      } 
      else { 
       difference -= 1; 
      } 
     } 
     else { 
      difference -= 1; 
     } 
    } 
    if (comparedDate.year > this.year) { 
     difference = comparedDate.year - this.year; 

     if (comparedDate.month > this.month) { 
     } 
     else if (comparedDate.month == this.month) { 
      if (comparedDate.day >= this.day) { 
      } 
      else { 
       difference -= 1; 
      } 
     } 
     else { 
      difference -= 1; 
     } 
    } 

    return difference; 
} 

接下来，让我们摆脱那些空分支。

public int differenceInYears(MyDate comparedDate) { 
    int difference; 

    if (this.year > comparedDate.year) { 
     difference = this.year - comparedDate.year; 

     if (this.month == comparedDate.month) { 
      if (this.day < comparedDate.day) { 
       difference -= 1; 
      } 
     } 
     else if (this.month < comparedDate.month) { 
      difference -= 1; 
     } 
    } 
    if (comparedDate.year > this.year) { 
     difference = comparedDate.year - this.year; 

     if (comparedDate.month == this.month) { 
      if (comparedDate.day < this.day) { 
       difference -= 1; 
      } 
     } 
     else if (comparedDate.month < this.month) { 
      difference -= 1; 
     } 
    } 

    return difference; 
} 

现在让我们看看，如果我们不能&&和||挤在一起的一些条件。

public int differenceInYears(MyDate comparedDate) { 
    int difference; 

    if (this.year > comparedDate.year) { 
     difference = this.year - comparedDate.year; 

     if (this.month == comparedDate.month && this.day < comparedDate.day || 
      this.month < comparedDate.month) 
     { 
      difference -= 1; 
     } 
    } 
    if (comparedDate.year > this.year) { 
     difference = comparedDate.year - this.year; 

     if (comparedDate.month == this.month && comparedDate.day < this.day || 
      comparedDate.month < this.month) 
     { 
      difference -= 1; 
     } 
    } 

    return difference; 
} 

这两个块看起来非常相似，不是吗？我们可以通过有条件地交换this和comparedDate来合并它们。假设a和b分别为更早和更晚的日期。

public int differenceInYears(MyDate comparedDate) { 
    MyDate a = (this.year < comparedDate.year) ? this : comparedDate; 
    MyDate b = (this.year < comparedDate.year) ? comparedDate : this; 

    int difference = b.year - a.year; 

    if (a.year < b.year) { 
     if (a.month == b.month && a.day < b.day || 
      a.month < b.month) 
     { 
      difference -= 1; 
     } 
    } 

    return difference; 
} 

最后一个挤压。

public int differenceInYears(MyDate comparedDate) { 
    MyDate a = (this.year < comparedDate.year) ? this : comparedDate; 
    MyDate b = (this.year < comparedDate.year) ? comparedDate : this; 

    int difference = b.year - a.year; 

    if (a.year < b.year && 
     (a.month == b.month && a.day < b.day || 
     a.month < b.month)) 
    { 
     difference -= 1; 
    } 

    return difference; 
} 

Answer 2

这里的关键是：有时需要“复杂”的计算。

从这个意义上说：不要担心效率低下。你应该花时间去编写真正可读的代码;非常像约翰的回答引导你。一般来说，您会将原则视为例如“single layer of abstraction”原则;以避免有这样冗长的if/else级联。

另一个关键点：这样的任务是完美单元测试。换句话说：您应该从编写一些简单的测试用例开始，这些测试用例只需使用各种（预先选择的）输入日期来调用您的方法;然后检查是否返回了预期结果。

换句话说：

你先专注于编写大量的测试用例，你能不能请你确定的你的代码是正确的......因为它使您能够重构你的代码，而不必担心打破吧。

使用JUnit是一样简单：

@Test 
public testDifferenceIs0 { 
    MyDate thisYear = new MyDate(1,1,2017); 
    MyDate thisYearToo = new MyDate(1,1,2017); 
    assertThat(thisYear.differenceInYears(thisYearToo), is(0)); 
} 

例如。

Answer 3

好了，所以一些考虑后，我认为这可能看起来更好一点：

public int differenceInYears(MyDate comparedDate){ 
    //Calculate days total 
    long daysTotalThisDate = this.year * 365 + this.month * 30 + this.day; 
    long daysTotalComparedDate = comparedDate.year * 365 + comparedDate.month * 30 + comparedDate.day; 

    //Get absolute value 
    long differenceInDays = daysTotalThisDate - daysTotalComparedDate; 
    if (differenceInDays < 0){ 
     differenceInDays *= -1; 
    } 

    //the (int) cast will always round down, so anything under 365 will be 0 
    return (int) differenceInDays/365; 
} 

这并没有考虑到闰年。

之前人们认为我的计算daysTotal...是错误的。你是对的。但是我对两次计算都做错了，所以最终结果还是可以的，因为我们只需要计算一年中的差异，而不是几天。