二维数组列表的排列 [英] Permutation of a 2 dimensional arraylist

问题描述

我正在尝试制作一个二维数组列表，该列表中填充了递归的每个可能的组合，比如 1,2,3,4.没有加倍.

I'm trying to make a 2 dimensional array list that is filled with every possible combination of lets say 1,2,3,4 recursively. with no double ups.

例如.
1,0,0
2,0,0
3,0,0
4,0,0
1,2,0
1,3,0
1,4,0​​
1,2,3
等等...

for example.
1,0,0
2,0,0
3,0,0
4,0,0
1,2,0
1,3,0
1,4,0
1,2,3
etc...

到目前为止我有

//this gives me all my numbers
for(int i =0;i<arraySize;i++)
index[i] = i;

// and is the part that make the combinations
for(int i = 0;i<arraySize;i++){
   for(int x = 0;x<k;x++)
      combinations.get(i).set(x, index[i]);

我也不想打印结果我想将结果存储在二维数组中

Also I'm not trying to print the result I want to store the result in a 2 dimensional array

推荐答案

另一个非递归选项.我用词"表示(部分)排列，用符号"表示数字:

Another, non-recursive option. I use "words" for (partial) permutations and "symbols" for numbers:

/**
 * Expands the set of existing words by adding non-repeated symbols to their right
 * @param symbols to choose from
 * @param existing words (may include the empty word)
 * @param expanded result 
 */
public static void expand(ArrayList<Integer> symbols, 
        ArrayList<ArrayList<Integer>> existing, 
        ArrayList<ArrayList<Integer>> expanded) {
    for (ArrayList<Integer> prev : existing) {
        Integer last = prev.isEmpty() ? null : prev.get(prev.size()-1);
        for (Integer s : symbols) {
            if (last == null || s > last) {
                ArrayList<Integer> toAdd = new ArrayList<>(prev);
                toAdd.add(s);
                expanded.add(toAdd);
            }
        }
    }
}

public static void main(String[] args) {

    ArrayList<Integer> symbols = new ArrayList<>(
            Arrays.asList(new Integer[]{1, 2, 3, 4}));

    ArrayList<ArrayList<Integer>> prev = new ArrayList<>();
    ArrayList<ArrayList<Integer>> next = new ArrayList<>();
    ArrayList<ArrayList<Integer>> aux;
    ArrayList<ArrayList<Integer>> output = new ArrayList<>();
    // add empty
    prev.add(new ArrayList<Integer>());
    // expand empty, then expand that expansion, and so on and so forth
    for (int i=0; i<symbols.size(); i++) {
        expand(symbols, prev, next);
        output.addAll(next);
        aux = prev; prev = next; next = aux;
        next.clear();
    }
    // print result (note: only for debugging purposes)
    for (ArrayList<Integer> o : output) {
        for (int i : o) System.out.print(i);  System.out.println();
    }
}

并且输出与原始问题匹配(这不是似乎要求实际排列，至少根据提供的示例):

And the output matches the original question (which does not seem to be asking for actual permutations, at least according to the provided example):

1
2
3
4
12
13
14
23
24
34
123
124
134
234
1234

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