Recursive Explanation (iv)

1  Public classA {2     Static intK=0;3     Static intTemp_k=0;4     Static intP=0;5     Static intQ=0;6      Public Static voidMain (String args[]) {7         Char[] ss = {' 1 ', ' 2 ', ' 3 ', ' 4 ', ' 5 ', ' 1 ', ' 2 ', ' 3 ', ' 4 ', ' 5 '};8Permutation (SS, 0);9         if(Temp_k = =k) {Ten System.out.println (k); One System.out.println (p); A System.out.println (q); -         } -     } the      Public Static voidPermutation (Char[] SS,inti) { -          -         if(ss = =NULL|| I < 0 | | i >ss.length) { -             return; +         } -         if(i = =ss.length) { +String A =NewString (ss);  Ak++; atTemp_k =K; -             if(k==3628800) - System.out.println (a); -}Else { -              for(intj = i; J < Ss.length; J + +) { -                 Chartemp = Ss[j];//swaps the prefix so that it produces the next prefix inSS[J] =Ss[i]; -Ss[i] =temp; top++; +Permutation (ss, I + 1); -q++; thetemp = Ss[j];//swap the prefixes back to continue with the previous prefix arrangement. *SS[J] =Ss[i]; $Ss[i] =temp;Panax Notoginseng             } -         } the     } +}

The number of 10 numbers arranged, K storage total arrangement, p, q are stored in the stack and the number of times out of the stack. By the way, the final arrangement is printed, the result:

5123451234362880098641009864100

The total number of permutations is 10!
The calculation law is as follows:

N	1	2	3	4	5	6				10
K	1	2	6	24	120	720				10!
P	1	4	15	64	325	1956				9864100

The visible algorithm is less efficient, the number of recursive stacks is much larger than the number of full permutations, and with the increase of n value, p/n tends to approximate the natural logarithm e.

Recursive Explanation (iv)