Question

我正在尝试寻找一种最佳方法，以便在其中包含映射的2个arrayLists中找到不同的元素。

例如a1和a2是其中的数组列表

a1 = [{“ val”：“ 1”，“ id”：“ 19”}，{“ val”：“ 2”，“ id”：“ 22”}，{“ val”：“ 3” ，“ id”：“ 2”}，{“ val”：“ 4”，“ id”：“ 49”}]]
a2 = [{“ val”：“ 1”，“ id”：“ 12”}，{“ val”：“ 2”，“ id”：“ 22”}，{“ val”：“ 3” ，“ id”：“ 32”}，{“ val”：“ 5”，“ id”：“ 52”}]]

预期输出为：

lOld = [{“ val”：“ 5”，“ id”：“ 52”}]
lNew = [{“ val”：“ 4”，“ id”：“ 49”}]

我对这个问题的解决方案是：

List<Map<String, String>> lOld = new ArrayList<Map<String,String>>();
for (int i = 0; i < a2.size(); i++) {
    boolean found = false;
    for (int j = 0; j < a1.size(); j++) {
        if(a2.get(i).get("val").equals(a1.get(j).get("val"))){
            found = true;
            break;
        }
    }
    if(found == false){
        lOld.add(a2.get(i));
    }
}

List<Map<String, String>> lNew = new ArrayList<Map<String,String>>();
for (int i = 0; i < a1.size(); i++) {
    boolean found = false;
    for (int j = 0; j < a2.size(); j++) {
        if(a1.get(i).get("val").equals(a2.get(j).get("val"))){
            found = true;
            break;
        }
    }
    if(found == false){
        lNew.add(a1.get(i));
    }
}

是否有解决此问题的最佳方法？

注意：arrayList内部的映射包含一个以上的值。仅以a1和a2为例。

Answer 1

将ArrayList转换为HashSet
- 这将为set.contains（）提供O（1）
使用Map.equals（）检查2个地图是否具有相同的键值对（即检查内容是否相同）

现在，您可以遍历一个集合并检查该元素是否存在于另一个集合中。这将为您提供线性时间，而不是二次方

Answer 2

解决此问题的步骤：

从arrayList a1合并映射到lOld，从arrayList a2合并映射到lNew
从lOld Map和lNew Map中获取密钥集
将两个键集中的公用键添加到临时arrayList
从两个键集中删除此临时arrayList。
如果要以arrayList形式输出，请将lOld和lNew映射添加到新创建的arrayLists

demo on Ideone

        // start- get arrayList a1 index entries, fetch map,put all maps into one map to 
        //remove duplicate keys,doesn't matter value- because they will be removed anyways)
        Map<String, String> lOld = new HashMap<String, String>();
        for (i = 0; i < a1.size(); i++) {
            HashMap<String, String> a1Map = a1.get(i);
            lOld.putAll(a1Map);
        }
        //end

        // start- get arrayList a2 index entries, fetch map,put all maps into other map to remove
        // duplicate keys,doesn't matter value- because they will be removed anyways)

        HashMap<String, String> lNew = new HashMap<String, String>();
        for (j = 0; j < a2.size(); j++) {
            HashMap<String, String> a2Map = a2.get(j);
            lNew.putAll(a2Map);
        }
       //end

        // check if first map keys (set) is in second map keys (set). 
       //if yes, add them into a list.
        List<String> toRemove = new ArrayList<>();
        Set<String> oldKeys = lOld.keySet();
        Set<String> newKeys = lNew.keySet();
        for (String oldKey : oldKeys) {
            if (lNew.containsKey(oldKey)) {
                toRemove.add(oldKey);
            }
        }

        // remove that list elements from both sets which will remove them from map itself.
        oldKeys.removeAll(toRemove);
        newKeys.removeAll(toRemove);

        // print both map
        System.out.println("lold map is: " + lOld);
        System.out.println("lNew map is: " + lNew);

        // don't remove elements from set while iterating. it will give ConcurrentModificationException

Answer 3

在HashSet中查找集合中项目的最佳性能。

在这里查看： https://www.baeldung.com/java-hashset-arraylist-contains-performance

有什么最佳的方法来查找2 arrayList中的不同的值

3 个答案: