[Leetcode] Permutation Sequence

Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and K, return the kth permutation sequence.

Note:given N would be between 1 and 9 inclusive.

The law of this string is a small to large sequence of numbers, the K-string is the K-large number sequence string, then the brute force traversal is feasible, but too slow.

Another method is Cantor unfold, time complexity O (n), Space complexity O (1).

1 classSolution {2  Public:3     stringGetpermutation (intNintk) {4vector<int>numbers;5          for(intI=1; i<=n;i++)6         {7 Numbers.push_back (i);8         }9vector<int>result;Ten         intb=k-1; One          for(inti=n-1; i>=0; i--) A         { -             inta=b/(Jiecheng (i)); - Result.push_back (Numbers[a]); theNumbers.erase (Numbers.begin () +a); -             if(i!=0) -             { -b=b%(Jiecheng (i)); +             } -         } +         stringResult_str; A          for(intI=0; I<result.size (); i++) at         { -             Charch=result[i]+ -; -result_str+=ch; -         } -         returnResult_str; -     } in     intJiecheng (intN) -     { to         intresult=1; +          for(intI=1; i<=n;i++) -         { theresult*=i; *         } $         returnresult;Panax Notoginseng     } -};

[Leetcode] Permutation Sequence