java在获取最高值和最低值时统一代码

Question

我的IDE给了我重复代码的警告，现在我试图找出有没有办法统一或找到这两种方法的共同抽象，这两种方法在结构上非常相似，但做的事情却相反？我自己想不出来。Thx

public double getLowestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    double lowestRate = Double.MAX_VALUE;
    for(Map<String, Double> ratesByDate : rates.values()){
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if(baseCurrecyRateToTargetCurrency < lowestRate){
            lowestRate = baseCurrecyRateToTargetCurrency;
        }
    }
    return lowestRate;
}

public double getHighestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    double highestRate = Double.MIN_VALUE;
    for(Map<String, Double> ratesByDate : rates.values()){
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if(baseCurrecyRateToTargetCurrency > highestRate){
            highestRate = baseCurrecyRateToTargetCurrency;
        }
    }
    return highestRate;
}

Answer 1

我同意前面的几个答案，并且相信它们可以很好地被使用

就我个人而言，我会创建一个统一的方法来处理公共逻辑流，然后有两个描述性方法称之为“公共逻辑方法”

因此，我不必将布尔标志作为参数处理（在本例中为双标志），我有机会表示哪个名称最适合该方法，最后我将所有逻辑流集中在一个方法中

示例：

public double getLowestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    return commonLogic(targetCurrency, getAllRatesInCurrentMonth(baseCurrency), Double.MAX_VALUE);
}

public double getHighestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency) {
    return commonLogic(targetCurrency, getAllRatesInCurrentMonth(baseCurrency), Double.MIN_VALUE);
}

private double commonLogic(String targetCurrency, Map<String, Map<String, Double>> rates, double rate) {
    for(Map<String, Double> ratesByDate : rates.values()){
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if(rate == Double.MIN_VALUE ? baseCurrecyRateToTargetCurrency > rate : baseCurrecyRateToTargetCurrency < rate){
            rate = baseCurrecyRateToTargetCurrency;
        }
    }
    return rate;
}

Answer 2

一个简单的解决方案如下： 给函数一个rate，它是double.MIN_VALUE或double.MAX_VALUE，并将其称为rate。在函数中，您可以检查速率是double.MIN_VALUE还是double.MAX_VALUE。或者只需检查rate是更小还是更大baseCurrecyRateToTargetCurrency。我不太确定它是常数还是非常数。这将是一个例子：

public double getCurrencyRateExtremaInTheCurrentMonth(String baseCurrency, String targetCurrency, double rate) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    double resultRate = rate
    for(String key : rates.keySet()){
        Map<String, Double> ratesByDate = rates.get(key);
        double baseCurrecyRateToTargetCurrency = ratesByDate.get(targetCurrency);
        if((rate == double.MIN_VALUE && baseCurrecyRateToTargetCurrency > rate) || (rate == double.MAX_VALUE && baseCurrecyRateToTargetCurrency < rate)){
            resultRate = baseCurrecyRateToTargetCurrency;
        }
    }
    return resultRate;
}

我假设baseCurrecyRateToTargetCurrency的值与在函数中以相同的方式检索的值相同。是这样吗

Answer 3

在我看来，里面有很多不必要的代码。使用流可以更好地实现同样的效果。看起来是这样的：

public double getHighestCurrencyRateInTheCurrentMonth(String baseCurrency, String targetCurrency, String variant) {
    Map<String, Map<String, Double>> rates = getAllRatesInCurrentMonth(baseCurrency);
    DoubleStream doubleStream = rates.values().stream()
                 .mapToDouble(entry -> entry.get(targetCurrency));
    OptionalDouble result = variant.equals("max") ? doubleStream.max() : doubleStream.min();
    return result.orElse(0);
}