最佳排序方式是根据地图中整数索引的列表进行比较

Question

我想对一个由整数组成的数组进行排序，这些整数是map的索引>，比较取决于Map中与整数索引相对应的字符串列表， 那么考虑到时间和内存，如果数组很大，那么哪种排序算法是最好的呢？

public class AuthorSorting {
    public static void main(String args[]) throws Exception{
        int index = 0;
        int array[] = {0, 1 ,2 ,3 ,4 ,5 ,6};
        Map<Integer,List<String>> myMap = new LinkedHashMap<>();
        myMap.put(index++, Arrays.asList("Abhijit Bhaduri","Kewal Dheer","Hemant Shesh"));
        myMap.put(index++, Arrays.asList("Kewal Dheer","Kiran Desai","Nandini Sahu"));
        myMap.put(index++, Arrays.asList("Abhijit Bhaduri"));
        myMap.put(index++, Arrays.asList("Kewal Dheer","Kiran Desai","Mohan Deep"));
        myMap.put(index++, Arrays.asList("Kewal Dheer"));
        myMap.put(index++, Arrays.asList("Kiran Desai","Mohan Deep","Pawan Karan"));
        myMap.put(index++, Arrays.asList("Hemant Shesh"));
        System.out.println("*****input*****\n");
        for(Map.Entry<Integer, List<String>> entry: myMap.entrySet()) {
            System.out.println(entry.getKey()+" "+entry.getValue().toString());
        }
        int temp;
        for(int i = 1 ; i < array.length;i++) {
            for(int j = 0; j < array.length-i;j++) {
                    if(compare(myMap.get(array[j]),myMap.get(array[j+1]))<0) {//swapping only if compare() returns -1
                        temp = array[j];
                        array[j] = array[j+1];
                        array[j+1] = temp;
                    }                   
            }
        }

        System.out.println("\n******result*******\n");
        for(int number:array) {
            System.out.println(number+" "+myMap.get(number).toString());
        }
    }
    static int compare(List<String> listOne,List<String> listTwo) {
        int i;
        for(i = 0 ; i<listOne.size()&&i<listTwo.size();i++) {
            if(listOne.get(i).compareToIgnoreCase(listTwo.get(i))>0) {
                return -1;// listOne's string is lexically greater
            }
            else if(listOne.get(i).compareToIgnoreCase(listTwo.get(i))<0) {
                return 1;// listTwo's string is lexically greater
            }
        }
        if(listOne.size()==listTwo.size())
            return 0;
        if(i==listOne.size())
            return 1;
        if(i==listTwo.size())
            return -1;
        return 0;
    }
} 

*****input*****

0 [Abhijit Bhaduri, Kewal Dheer, Hemant Shesh]
1 [Kewal Dheer, Kiran Desai, Nandini Sahu]
2 [Abhijit Bhaduri]
3 [Kewal Dheer, Kiran Desai, Mohan Deep]
4 [Kewal Dheer]
5 [Kiran Desai, Mohan Deep, Pawan Karan]
6 [Hemant Shesh]

******result*******

2 [Abhijit Bhaduri]
0 [Abhijit Bhaduri, Kewal Dheer, Hemant Shesh]
6 [Hemant Shesh]
4 [Kewal Dheer]
3 [Kewal Dheer, Kiran Desai, Mohan Deep]
1 [Kewal Dheer, Kiran Desai, Nandini Sahu]
5 [Kiran Desai, Mohan Deep, Pawan Karan]

当前，我已经使用了冒泡排序，还不确定是否可以在大型数组中使用