允许重复的TreeSet或TreeMap [英] A TreeSet or TreeMap that allow duplicates

问题描述

我需要一个集合排序元素，但不会删除重复项。

I need a Collection that sorts the element, but does not removes the duplicates.

TreeSet 。

c $ c>实际上将值添加到支持的 TreeMap ：

code> public boolean add（E e）{
return m.put（e，PRESENT）== null;
}

public boolean add(E e) { return m.put(e, PRESENT)==null; }

TreeMap使用 / code> 比较逻辑

c $ c> Comparator 在等于元素的情况下返回1而不是0。
因此在等于元素的情况下， TreeSet 与此 Comparator 将不会覆盖重复，

I have written a Comparator that returns 1 instead of 0 in case of equal elements. Hence in the case of equal elements the TreeSet with this Comparator will not overwrite the duplicate and will just sort it.

我已经测试过简单的 String 对象，但我需要一组自定义对象。 / p>

I have tested it for simple String objects, but I need a Set of Custom objects.

public static void main(String[] args)
{       
        List<String> strList = Arrays.asList( new String[]{"d","b","c","z","s","b","d","a"} );      
        Set<String> strSet = new TreeSet<String>(new StringComparator());       
        strSet.addAll(strList);     
        System.out.println(strSet); 
}

class StringComparator implements Comparator<String>
{
    @Override
    public int compare(String s1, String s2)
    {
        if(s1.compareTo(s2) == 0){
            return 1;
        }
        else{
            return s1.compareTo(s2);
        }
    }
}

有更好的方法来实现这个？

Is this approach fine or is there a better way to achieve this?

EDIT

以下类的ArrayList：

Actually I am having a ArrayList of the following class:

class Fund 
{
    String fundCode;
    BigDecimal fundValue;
    .....

    public boolean equals(Object obj) {
    // uses fundCode for equality
    }
}

我需要所有 fundCode fundValue

I need all the fundCode with highest fundValue

推荐答案

我需要所有fundCode具有最高fundValue

I need all the fundCode with highest fundValue

如果这是你想要排序的唯一原因，我会建议不要排序。排序主要具有 O（n log（n））的复杂性。找到最大值只有一个复杂的 O（n），并在您的列表上的一个简单的迭代实现：

If that's the only reason why you want to sort I would recommend not to sort at all. Sorting comes mostly with a complexity of O(n log(n)). Finding the maximum has only a complexity of O(n) and is implemented in a simple iteration over your list:

List<Fund> maxFunds = new ArrayList<Fund>();
int max = 0;
for (Fund fund : funds) {
    if (fund.getFundValue() > max) {
        maxFunds.clear();
        max = fund.getFundValue();

    }
    if (fund.getFundValue() == max) {
        maxFunds.add(fund);

    }
}

第三级库，如 Guava 。请参阅：如何从List中获取max（）元素

You can avoid that code by using a third level library like Guava. See: How to get max() element from List in Guava

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