The Fake Geek's blog: Find possible triangle triplets

The Fake Geek's blog: Find possible triangle triplets
"Given a array of positive integers, find all possible triangle triplets that can be formed from this array. 
eg: 9 8 10 7 
ans: 9 8 10, 9 8 7, 9 10 7, 7 8 10 
Note : array not sorted, there is no limit on the array length"
http://www.careercup.com/question?id=5863307617501184

 public void triangleTriplet(int a[])
 {
  int n=a.length;
  for(int i=0;i<n;i++)
  {
   for(int j=0;j<n;j++)
   {
    for(int k=0;j<n;j++)
    {
     if(i!=j && j!=k && i!=k)
     if(a[i]+a[j]>a[k]  && a[j]+a[k]>a[i] && a[i]+a[k]>a[j])
     {
      System.out.println(a[i]+" "+a[j]+" "+a[k]);
     }
    }
   }
  }
 }

I used backtracking for this problem. However, it may not be the best solution since the complexity is O(n!).
?????

 public ArrayList<ArrayList <Integer>> possibleTriangle(int[] nums)

 {

  ArrayList<ArrayList <Integer>> rst = new ArrayList<>();

  if (nums == null || nums.length == 0)

   return rst;

  ArrayList<integer> triplets = new ArrayList<integer>();

  Arrays.sort(nums);

  getTriangle(rst, triplets, nums, 0);

  return rst;

 }

 private void getTriangle(ArrayList<ArrayList <Integer>> rst, ArrayList<integer> triplets,

   int[] nums, int start)

 {

  if (triplets.size() == 3 && triplets.get(0) + triplets.get(1) > triplets.get(2))

   rst.add(new ArrayList<integer> (triplets));

  for (int i = start; i < nums.length; i++)

  {

   if (i != start && nums[i] == nums[i - 1])

    continue;

   triplets.add(nums[i]);

   getTriangle(rst, triplets, nums, i + 1);

   triplets.remove(triplets.size() - 1);

  }

 }

 public void printList(ArrayList<arraylist nteger="">> rst)

 {

  for (int i = 0; i < rst.size(); i++)

   System.out.println(rst.get(i));

 }

 

}

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