递归生成列表的所有可能排列

问题描述

我试图递归地生成列表中的所有项目.我已经看到了一些类似问题的解决方案，但我无法让我的代码工作.有人可以指出我如何修复我的代码吗?

I'm trying to recursively generate all items in a list recursively. I've seen a few solutions to similar questions to this, but I haven't been able to get my code to work. Could someone point out how I can fix my code?

这对所有 S/O'ers 开放，而不仅仅是 Java 人.

This is open to all S/O'ers, not just Java people.

(另外我应该注意到它会因 SO 异常而崩溃).

(Also I should note that it crashes with a SO exception).

样本输入:

[1, 2, 3]

输出:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

//allPossibleItems is an AL of all items

//this is called with generatePerm(null, new ArrayList<Item>);

private void generatePerm(Item i, ArrayList<Item> a) {
    if (i != null) { a.add(i); }
    if (a.size() == DESIRED_SIZE) {
        permutations.add(a);
        return;
    }
    for (int j = 0; j < allPossibleItems.size(); j++) {
        if (allPossibleItems.get(j) != i)
            generatePerm(allPossibleItems.get(j), a);
    }
}

推荐答案

如果 allPossibleItems 包含两个不同的元素，x 和 y，那么您依次将 x 和 y 写入列表，直到到达 DESIRED_SIZE.这真的是你想要的吗?如果您选择的 DESIRED_SIZE 足够大，您将在堆栈上有太多递归调用，因此出现 SO 异常.

If allPossibleItems contains two different elements, x and y, then you successively write x and y to the list until it reaches DESIRED_SIZE. Is that what you really want? If you pick DESIRED_SIZE sufficiently large, you will have too many recursive calls on the stack, hence the SO exception.

我会做什么(如果原件没有双胞胎/重复项)是:

What I'd do (if original has no douplets / duplicates) is:

public <E> List<List<E>> generatePerm(List<E> original) {
  if (original.isEmpty()) {
    List<List<E>> result = new ArrayList<>();
    result.add(new ArrayList<>());
    return result;
  }
  E firstElement = original.remove(0);
  List<List<E>> returnValue = new ArrayList<>();
  List<List<E>> permutations = generatePerm(original);
  for (List<E> smallerPermutated : permutations) {
    for (int index = 0; index <= smallerPermutated.size(); index++) {
      List<E> temp = new ArrayList<>(smallerPermutated);
      temp.add(index, firstElement);
      returnValue.add(temp);
    }
  }
  return returnValue;
}

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