如何只返回出现次数最少的字符串ArrayList？

Question

我有一个String[],originalStringArray其中有重复的.所以{"dog","cat","dog","fish","dog","cat"}.

我想创建一个只返回恰好发生一定次数的字符串的函数.在这里,如果我说3,它将返回"狗"而不是"猫".

这是我目前的代码:

public ArrayList<String>  returnMultiples(String[] originalStringArray,int requiredCount){
    ArrayList<Integer> mCount = new ArrayList<>();
    List<String> list = Arrays.asList(originalStringArray);
    ArrayList<String> result = new ArrayList<>();

    // Count occurrences in original string
    for(String item: originalStringArray){
        mCount.add(Collections.frequency(list,item));
    }

    // If frequency is equal to count, add to array list
    for(int i=0; i<mCount.size(); i++){
        if(mCount.get(i) == requiredCount){
            result.add(originalStringArray[i]);
        }
    }

    return result;
}

我遇到的问题是,我在某处读到了Collections库非常慢并且拖动,并且似乎可以使用HashSets和表来减少此方法.不幸的是,我对如何做到这一点感到很茫然.有一个更好的方法吗？

Answer 1

需要某种地图来执行此操作。这是使用 HashMap 编写的示例：

public ArrayList<String> returnMultiples(String[] array, int min){
    HashMap<String, Integer> counts = new HashMap<String, Integer>();//instantiate a new HashMap

    //loop through the array and count the occurrences of each different string in the array
    for(int i = 0; i < array.length; i++){
        String word = array[i];
        if(counts.containsKey(word))
            counts.put(word, counts.get(word) + 1);
        else
            counts.put(word, 1);
    }

    ArrayList<String> multiples = new ArrayList<String>();

    //check if any of the words occur >= min times. if so, add them to the returning list.
    for(String key : counts.keySet()){
        if(counts.get(key) >= min){
            multiples.add(key);
        }
    }

    return multiples;//return the list we just created of the desired strings
}

根据字符串的长度，HashMap 比使用集合更有效，尽管差异几乎可以忽略不计。