Permutation Sequence Java Solutions

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.

1  Public classSolution {2      PublicString Getpermutation (intNintk) {3StringBuilder ans =NewStringBuilder ();4StringBuilder tmp =NewStringBuilder ();5          for(inti = 1; i<=n; i++) tmp.append (i);6         intFactor =calculate (n);7         intindex = 0;8          for(inti = n; i>=1; i--){9Factor/=i;Tenindex = (k-1)/factor; One ans.append (Tmp.charat (index)); AK-= index*factor; - Tmp.deletecharat (index); -         } the         returnans.tostring (); -     } -      -      Public intCalculateintN) { +         intsum = 1; -          for(;n>=1; n--){ +Sum *=N; A         } at         returnsum; -     } -}

N Number of permutation total has n factorial, based on this property we can get a corresponding number of which is. The idea is that, for example, the current length is n, and we know that each of the same starting elements corresponds (N-1)! A permutation,

That is (n-1)! After a permutation, a starting element is changed. Therefore, as long as the current k is carried out (N-1)! The resulting number is the index of the current remaining array so that the corresponding element can be obtained.

Https://discuss.leetcode.com/topic/17348/explain-like-i-m-five-java-solution-in-o-n

Permutation Sequence Java Solutions