由网友(蝴蝶少女。)分享简介：我想递归生成列表中的所有项目递归。我已经看到了一些解决方案，以类似的问题，但我一直没能得到我的code工作。可能有人指出，如何可以解决我的code？I'm trying to recursively generate all items in a list recursively. I've seen a few s...

我想递归生成列表中的所有项目递归。我已经看到了一些解决方案，以类似的问题，但我一直没能得到我的code工作。可能有人指出，如何可以解决我的code？

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?

这是开放给所有的S / O'ers，而不仅仅是Java的人。

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

（我也应该注意，与SO异常崩溃）。

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

样品输入： [1，2，3]

Sample input: [1, 2, 3]

输出： [1，2，3] [1，3,2] [2，1,3] [2，3，1] [3，1,2] [3,2，1]

Output: [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包含两种不同的元素，x和y，则依次写入x和y以列表中，直到它达到DESIRED_SIZE。这是你真正想要的？如果你选择DESIRED_SIZE足够大，你将不得不在堆栈上太多的递归调用，因此SO例外。

If allPossibleItems contains two different elements, x and y, then you successively write x and y to the list a 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（如果原来没有douplets）：

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

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