Question

这是我到目前为止的代码，尽管数组的大小需要更改。我需要使用循环向数组添加值;要做到这一点，我想我需要使用嵌套的ArrayList。但是，我能否以与下面已经完成的方式相同的方式对ArrayList进行排序，还是有一种方法可以添加到这个数组中？

 public static void main(final String[] args) {

     String[][] data = new String[][] {
            new String[] { "Roy","3", "12" },
            new String[] { "Daniel","5", "22" },
            new String[] { "Joe","2", "3" },
            new String[] { "Ryan","4", "4" },
            new String[] { "Leroy","2", "5" },
            new String[] { "Smith","1", "6" },};



    Arrays.sort(data, new Comparator<String[]>() {
        @Override
        public int compare(String[] entry1, final String[] entry2) {
             String st1 = entry1[2];
             String st2 = entry2[2];
             int n1 = Integer.parseInt(st1);
             int n2 = Integer.parseInt(st2);
             return Integer.compare(n2, n1);

        }
    });




    for ( String[] s : data) {
        System.out.println(s[0] + "            " + s[1]+ "           " + s[2]);
    }

}

Answer 1

是的，似乎没有问题。将String[][]的类型更改为data（数组 list 或字符串数组），而不是Arraylist<String[]>（字符串数组的数组）。所以这看起来像：

public static void main(final String[] args) {

     ArrayList<String[]> data = new ArrayList<String[]>();
     data.add(new String[] { "Roy","3", "12" });
     data.add(new String[] { "Daniel","5", "22" });
     data.add(new String[] { "Joe","2", "3" });
     data.add(new String[] { "Ryan","4", "4" });
     data.add(new String[] { "Leroy","2", "5" });
     data.add(new String[] { "Smith","1", "6" });

     Collections.sort(data, new Comparator<String[]>() {
        @Override
        public int compare(String[] entry1, final String[] entry2) {
             String st1 = entry1[2];
             String st2 = entry2[2];
             int n1 = Integer.parseInt(st1);
             int n2 = Integer.parseInt(st2);
             return Integer.compare(n2, n1);
        }
    });

    //Other stuff...
}

也许你的意思是你希望内部字符串数组是动态大小的。这也没有问题，它完全按照你的想法嵌套。

public static void main(final String[] args) {

     String[][] rawData = new String[][] {
        new String[] { "Roy","3", "12" },
        new String[] { "Daniel","5", "22" },
        new String[] { "Joe","2", "3" },
        new String[] { "Ryan","4", "4" },
        new String[] { "Leroy","2", "5" },
        new String[] { "Smith","1", "6" }};

     ArrayList<ArrayList<String>> data = new ArrayList<ArrayList<String>>();
     for(String[] arr : rawData){
         ArrayList<String> a = new ArrayList<String>();
         data.add(a);
         for(String s : arr){
             a.add(s);
         }
     }

     Collections.sort(data, new Comparator<ArrayList<String>>() {
        @Override
        public int compare(ArrayList<String> entry1, ArrayList<String> entry2) {
             String st1 = entry1.get(2);
             String st2 = entry2.get(2);
             int n1 = Integer.parseInt(st1);
             int n2 = Integer.parseInt(st2);
             return Integer.compare(n2, n1);
        }
    });

    //Other stuff...
}

在这两种情况下，您都应该注意比较器访问索引2。如果数组（或arraylist）的长度仅为1，会发生什么？在这种情况下，您希望比较器做什么？它们是有希望整数但可能不是整数的字符串列表。如果比较器试图将12与“hello”进行比较，你希望比较器做什么？

排序/创建嵌套的arraylists

1 个答案: