java提高了合并两个ArrayList的性能

Question

我正在用下面的代码合并两个ArrayList。代码正在运行，并给了我想要的结果，但我想要一个更高效的版本。以下是条件

1. 方法接受两个列表，两个列表中的元素按降序排列（5,4,3,2）
2. 方法接受一个整数来决定结果ArrayList的大小
3. 第一个输入列表的大小永远不会大于结果ArrayList的大小

代码：

public ArrayList<Integer> mergeList(ArrayList<Integer> first,ArrayList<Integer> second, int n){
    //case 1: when both list are null.
    if(first == null && second == null )
        return null;
    //case 2: when first list is null but second list have elements
    else if( first == null && second != null){
        return second.size() >=n ? new ArrayList<Integer>(second.subList(0, n)) : second;
    }
    //case 3: when first list have record and second list is null
    else if(first != null && second == null){
        return first;
    }
    //case 4: when both list have elements 
    else {
        first.addAll(second);
        Collections.sort(first);
        Collections.reverse(first);
        return first.size()>=n ? new ArrayList<Integer>(first.subList(0, n)) : first;
    }
}

}

Answer 1

看起来您试图保存first和second的内容。如果您没有，那么这将对您很好，并将使您的代码更快、更可读：

public ArrayList<Integer> mergeList(ArrayList<Integer> first,ArrayList<Integer> second, int maxLength){

    //case 1: when both list are null.
    if(first == null && second == null )
        return null;
    //case 2: when first list is null but second list have elements
    else if( first == null && second != null){
        return second;
    }
    //case 3: when first list have record and second list is null
    else if(first != null && second == null){
        return first;
    }
    //case 4: when both list have elements 
    else if(first != null && second != null){
        first.addAll(second);
        Collections.sort(first); //want to merge these two line into one
        Collections.reverse(first);
    }
    return (ArrayList) first.size() > maxLength ? first.subList(0, n) : first;
}

这之所以更快，是因为对于每个addAll()，Java必须迭代所有项，将它们复制到tempList。我保留了Collections.reverse调用，因为您似乎需要按相反的顺序对数据进行排序