1>>> from itertools import permutations
2>>> perms = [''.join(p) for p in permutations('stack')]
3>>> perms
4

1/* to be used something like this:
2int [] toBePermuted = new int [] {1, 2, 3, 4};
3ArrayList<int[]> a = heap(toBePermuted);
4any mention of int [] can be replaced with any other Array of objects */
5
6ArrayList<int []> heap(int [] input) {
7  ArrayList<int []> ret = new ArrayList<int []> ();
8  ret = generate(input.length, input, ret);
9  return ret;
10}
11
12ArrayList<int []> generate(int k, int [] a, ArrayList<int []> output) {
13  if (k == 1) {
14    output.add(a.clone());
15  } else {
16    output = generate(k-1, a, output);
17    for (int i=0; i<k-1; i++) {
18      if (k%2 == 0) {
19        int temp = a[i];
20        a[i] = a[k-1];
21        a[k-1] = temp;
22      } else {
23        int temp = a[0];
24        a[0] = a[k-1];
25        a[k-1] = temp;
26      }
27      generate(k-1, a, output);
28    }
29  }
30  return output;
31}

1void permute(string a, int l, int r)  
2{  
3    // Base case  
4    if (l == r)  
5        cout<<a<<endl;  
6    else
7    {  
8        // Permutations made  
9        for (int i = l; i <= r; i++)  
10        {  
11  
12            // Swapping done  
13            swap(a[l], a[i]);  
14  
15            // Recursion called  
16            permute(a, l+1, r);  
17  
18            //backtrack  
19            swap(a[l], a[i]);  
20        }  
21    }  
22}  

1void perm(char a[], int level){
2
3    static int flag[10] = {0};
4    static char res[10];
5    // If we are the last character of the input string 
6    if(a[level] == '\0'){
7        // First we assign stopping point to result
8        res[level] = '\0';
9        // Now we print everything
10        for(int i = 0; res[i] != '\0'; ++i){
11            printf("%c", res[i]);
12        }
13        printf("\n");
14        ++counter;
15    }
16    else{
17        // Scan the original string and flag to see what letters are available
18        for(int i = 0; a[i] != '\0'; ++i){
19            if(flag[i] == 0){
20                res[level] = a[i];
21                flag[i] = 1;
22                perm(a, level + 1);
23                flag[i] = 0;
24            }
25        }
26    }
27}
28
29int main(){
30    char first[] = "abc";
31    perm(first, 0);
32    return 0;
33}

1void permutation(string s)
2{
3    sort(s.begin(),s.end());
4	do{
5		cout << s << " ";
6	}
7    while(next_permutation(s.begin(),s.end()); // std::next_permutation
8    
9    cout << endl;
10}

1void find_permutations(vector<int> &array, size_t index, vector<int> current_perm, vector<vector<int>> &res){
2    if(index == array.size()) 
3        res.push_back(current_perm);
4    else{
5        for(size_t i = 0; i <= current_perm.size(); ++i){
6            vector<int> new_perm(current_perm.begin(), current_perm.end());
7            new_perm.insert(new_perm.begin()+i, array[index]);
8            find_permutations(array, index+1, new_perm, res);
9        }
10    }
11}