[Lintcode] Permutations

permutations

Given A list of numbers, return all possible permutations.

Example

For nums = [1,2,3] , the permutations is:

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

Challenge

Do it without recursion.

Solution 1:

Recursive

Permutations problem, with subsets basically is similar, just in to the result in the list, not to join the subset but join the whole arrangement, so to add some conditions, List size equals nums size can join.

classSolution {/**     * @paramnums:a List of integers. * @return: A List of permutations. */     PublicArraylist<arraylist<integer>> Permute (arraylist<integer>nums) {ArrayList<ArrayList<Integer>> result =NewArraylist<arraylist<integer>>(); if(Nums = =NULL|| Nums.size () = = 0){            returnresult; } ArrayList<Integer> list =NewArraylist<integer>();        Helper (result, list, nums); returnresult; }    Private voidHelper (arraylist<arraylist<integer>> result, arraylist<integer> list, arraylist<integer>nums) {        if(list.size () = =nums.size ()) {Result.add (NewArraylist<integer>(list)); return; }         for(inti = 0; I < nums.size (); i++){            if(List.contains (Nums.get (i))) {Continue;            } list.add (Nums.get (i));            Helper (result, list, nums); List.remove (List.size ()+ W); }    }}

View Code

Solution 2:

Non-recursive

Non-recursive, using the nine chapters of the answer, the basic idea of using a stack to simulate the DFS process. Recite it, just.

classSolution {/**     * @paramnums:a List of integers. * @return: A List of permutations. */     PublicArraylist<arraylist<integer>> Permute (arraylist<integer>nums) {ArrayList<ArrayList<Integer>>permutations=NewArraylist<arraylist<integer>>(); if(Nums = =NULL|| Nums.size () = = 0) {            returnpermutations; }                intn =nums.size (); ArrayList<Integer> stack =NewArraylist<integer>(); Stack.add (-1);  while(Stack.size ()! = 0) {Integer last= Stack.get (Stack.size ()-1); Stack.remove (Stack.size ()-1); //increase The last number            intNext =-1;  for(inti = last + 1; I < n; i++) {                if(!stack.contains (i)) {Next=i;  Break; }            }            if(Next = =-1) {                Continue; }                        //generate the next permutationStack.add (next);  for(inti = 0; I < n; i++) {                if(!stack.contains (i))                {Stack.add (i); }            }                        //Copy to permutations setarraylist<integer> permutation =NewArraylist<integer>();  for(inti = 0; I < n; i++) {Permutation.add (Nums.get (Stack.get (i)));        } permutations.add (permutation); }                returnpermutations; }}

[Lintcode] Permutations