[LeetCode] [Java] Permutation Sequence

[LeetCode] [Java] Permutation Sequence
Question:

 

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 will be between 1 and 9 inclusive.

   Question:

   Set [1, 2, 3 ,..., N] contains a total of n! Different Combinations exist and are unique.

   List and mark all these combinations in order. We get the following sequence (when n = 3 ):

    

   1. 123
   2. 132
   3. 213
   4. 231
   5. 312
   6. 321Given n and k, the k Elements in the sequence are returned. N is a number between 1 and 9.

       

      Algorithm analysis:

      Method 1:

      Using the method of the question permutations, we first get all the arrays and then find the k elements, but it times out. The Code will also be provided later.

      Method 2:

      Reference blog http://www.cnblogs.com/springfor/p/3896201.html

       

      The mathematical solution is not suitable for this question. Let's take a look.

       

      The question tells us:N can be n! Sort; then there will be (n-1) n-1 numbers )! .

      For n-digit digits, if the highest bit is not included, the n-1 bit after it will have(N-1 )! .

      Therefore, for n-digits, each of the different highest digits can be spliced (n-1 )! .

      So you can think of it as following each group (n-1 )! Group.

      K/(n-1 )! You can obtain the index of the highest bit in the sequence. In this way, the highest bit in the k-th arrangement can be obtained from the index bit in the series, and this number must be deleted from the series.

      Then, the new k can have k % (n-1 )! . (Forgive me for being stupid. I don't want to know why I want to update k ~~)

      Loop n times. At the same time, in order to be able to match the array coordinates, k first --.

      
      AC code:

      Method 1: (timeout)

       

      Public String getPermutation (int n, int k) {int num [] = new int [n]; ArrayList> temres = new ArrayList> (); for (int I = 0; I
           
            
      > Permute (int [] num) {ArrayList> result = new ArrayList> (); permute (num, 0, result); return result ;} static void permute (int [] num, int start, ArrayList> result) {if (start> = num. length) // termination condition, recursively returning to the end node is to convert the array into a linked list {ArrayList
            
             
      Item = convertArrayToList (num); result. add (item) ;}for (int j = start; j <= num. length-1; j ++) {swap (num, start, j); // exchange permute (num, start + 1, result ); // recursive swap (num, start, j) of the child after switching; // restore to the initial state before switching to facilitate the next exchange result} private static ArrayList
             
              
      ConvertArrayToList (int [] num) // array variable linked list {ArrayList
              
               
      Item = new ArrayList
               
                
      (); For (int h = 0; h <num. length; h ++) item. add (num [h]); return item;} private static void swap (int [] a, int I, int j) // exchange {int temp = a [I]; a [I] = a [j]; a [j] = temp ;}
               
              
             
            
           

      
      Method 2:

       

       

      Public String getPermutation (int n, int k) {k --; // to transfer it as begin from 0 rather than 1 List
           
            
      NumList = new ArrayList
            
             
      (); For (int I = 1; I <= n; I ++) numList. add (I); int factorial = 1; for (int I = 2; I <n; I ++) factorial * = I; StringBuilder res = new StringBuilder (); // simple replacement of StringBuffer | used by a single thread in the string buffer | it is faster than StringBuffer (most of time) int times = n-1; while (times> = 0) {int indexInList = k/factorial; res. append (numList. get (indexInList); numList. remove (indexInList); k = k % factorial; // new k for next turn if (times! = 0) factorial = factorial/times; // new (n-1 )! Times --;} return res. toString ();}