如何将arrayList元素转换为2D数组？

I have an ArrayLists called List<Pair> patternList = new ArrayList<>() that contains a number of patterns of length 2. for example, for A={1,2,3,4} I create some patterns like (1,3)(3,2)...etc. I want to put these patterns into a 2d array(matrix)[A][A] in way that the patterns must go into a specific index in the matrix, For instance pattern (1,3) must go into index [3][3] or pattern (3,2) must go in [2][2].

我有一个名为List 的ArrayLists，其中包含多个长度为2的模式的listList =>（）。例如，对于A = {1,2,3,4}，我创建了一些模式，如（1， 3）（3,2）...等。我想把这些模式放到二维数组（矩阵）[A] [A]中，模式必须进入矩阵中的特定索引，例如模式（1,3）必须进入索引[3] [ 3]或模式（3,2）必须进入[2] [2]。

Thanks.

谢谢。

2 个解决方案

#1

If you just want to put them in by the second index. (edited to allow multiple per index):

如果你只想把它们放在第二个索引中。（编辑为允许每个索引多个）：

List<Pair>[][] matrix = new LinkedList<Pair>[5][5]; //Replace 5 here a getter for the maximum value of A
                                  //Here it's 5 because the max of your example is 4.
//Initialize all positions of matrix to empty list.
for (int r = 0; r < matrix.length; r++){
    for(int c = 0; c < matrix[r].length; c++){
        matrix[r][c] = new LinkedList<Pair>();
    }
}
for (Pair p : patternList){
    if (matrix[p.second][p.second] == null)
        matrix[p.second][p.second] = new LinkedList<Pair>();
    matrix[p.second][p.second].add(p);
}

This isn't really using the 2d aspect of the array though; you could accomplish exactly the same thing with a 1d array of lists:

这不是真正使用数组的第二方面;你可以使用一列列表完成同样的事情：

List<Pair>[] arr = new LinkedList<Pair>[5]; //Replace 5 here a getter for the maximum value of A
                                  //Here it's 5 because the max of your example is 4.
//Initialize all positions of array to empty list.
for (int r = 0; r < matrix.length; r++){
    arr[r] = new LinkedList<Pair>();
}
for (Pair p : patternList){
    if (arr[p.second] == null)
        arr[p.second] = new LinkedList<Pair>();
    arr[p.second].add(p);
}

#2

Since you want to keep more than one element at each position, I suggest you use a Multimap. A Multimap is basically a Map<Key, Collection<Value>> that takes care of creating the collections for you.

由于您希望在每个位置保留多个元素，我建议您使用Multimap。 Multimap基本上是Map >，负责为您创建集合。，collection>

Multimap

Multimap之

 Multimap<List<Integer>, Pair> myMap = ArrayListMultimap.create();

 for (Pair p : myPairs){
     List<Integer> key = Lists.newArrayList(p.second, p.second);
     myPair.put(key, p);
 }

#1