java - Java中使用Comparator对ArrayList进行排序

Question

我在按升序对 ArrayList 进行排序时遇到问题。我正在使用 Comparator 和 Collat​​or Of Collection 类。如何实现预期的排序顺序？非常感谢您的帮助。

算法计算出的升序为:

[AutomationRejectNotification|,AutomationRejectNotification1011, AutomationRejectNotification1021,AutomationTestNotification1, AutomationTestNotification100,AutomationTestNotification2,testDisplay Template, Testing Chrome, Testing Field, Test Notfication, testnotif, Test Notification #1]

预期的升序排序顺序是:

[AutomationRejectNotification1011, AutomationRejectNotification1021, AutomationRejectNotification|,AutomationTestNotification1, AutomationTestNotification2,AutomationTestNotification100,Test Notfication, Test Notification #1, testDisplay Template, Testing Chrome, Testing Field, testnotif]

Java代码:

public static void listSort(List<String> o1, boolean order) {
        final Pattern p = Pattern.compile("^\\d+");
        Comparator<String> c = new Comparator<String>() {
            public int compare(String object1, String object2) {
                Collator collator = Collator.getInstance(Locale.US);
                Matcher m = p.matcher(object1);

                if (!m.find()) {
                    return collator.compare(object1, object2);
                } else {
                    Long number2 = null;
                    Long number1 = Long.parseLong(m.group());

                    m = p.matcher(object2);
                    if (!m.find()) {
                        return collator.compare(object1, object2);
                    } else {

                        number2 = Long.parseLong(m.group());

                        int comparison = number1.compareTo(number2);
                        if (comparison != 0) {
                            return comparison;
                        } else {
                            return collator.compare(object1, object2);
                        }
                    }
                }


            }
        };
        o1.sort(c);

Answer 1

您的Comparator没有正确实现Comparator接口(interface)的约定，因此一切都失败了。 compare() 方法必须具有所有这些属性:

sgn(compare(x, y)) == -sgn(compare(y, x)) for all x and y.

((compare(x, y)>0) && (compare(y, z)>0)) implies compare(x, z)>0.

compare(x, y)==0 implies that sgn(compare(x, z))==sgn(compare(y, z)) for all z.

( API docs )

当要比较的两个字符串都包含非空十进制数字序列时，您仅通过数字进行比较，而如果至少一个字符串不包含任何数字，则您通过整理器进行比较。这会产生以下结果:

compare("z1", "y") > 0

compare("y", "x3") > 0

compare("z1", "x3") < 0

，不符合第二个必需属性(传递性)。

您可能想要做的是比较最长的前导无数字子字符串作为第一个标准，通过比较尾随数字(如果有)来打破任何联系。可能看起来像这样:

public static void listSort(List<String> o1, boolean order) {
    final Pattern p = Pattern.compile("([^0-9]*)([0-9]+)?");
    final Collator collator = Collator.getInstance(Locale.US);
    Comparator<String> c = new Comparator<String>() {
        public int compare(String object1, String object2) {
            Matcher m1 = p.matcher(object1);
            Matcher m2 = p.matcher(object2);

            if (!m1.lookingAt() || !m2.lookingAt()) {
                assert false : "Should never happen";
            }

            int result = collator.compare(m1.group(1), m2.group(1));

            if (result == 0) {
                String digits1 = m1.group(2);
                String digits2 = m2.group(2);

                if (digits1 != null && digits2 != null) {
                    Long number1 = Long.valueOf(digits1);
                    Long number2 = Long.valueOf(digits2);
                    result = number1.compareTo(number2);
                } else if (digits1 != null) {
                    result = 1;
                } else if (digits2 != null) {
                    result = -1;
                }
            }

            return result;
        }
    };
    o1.sort(c);
}

这将与您提供的预期顺序一致，但还有其他排序方案也会为这些特定元素产生相同的结果。