[Leetcode] 60. Permutation Sequence alignment 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 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.
• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

 

The meaning of problems

1 ~ n has n! Permutations, which are arranged to identify the k-th row of lexicographically.

 

Thinking

This is a very mathematical problem, what fancy algorithms do not, you have to go to the law of doom

 

 

code

 1  public String getPermutation(int n, int k) {
 2         char[] result = new char[n];
 3         List<Integer> list = new ArrayList<>();
 4         int[] factorial = new int[n];
 5 
 6         factorial[0] = 1;
 7         for (int i = 1; i < n; i++) {
 8             factorial[i] = factorial[i-1] *i;
 9         }
10         // factorial: 1, 1, 2
11          // denotes the number n corresponding to the full array, and 
12 is  
13 is          for ( int I =. 1; I <= n; I ++ ) {
 14              List.add (I);
 15          }
 16          // usable digital 1-2-3 , with the list because facilitate later delete digital 
. 17  
18 is          K--; // pick a high number
 19          // starting from the highest bit to the last bit to generate a result 
20 is          for ( int I = 0; I < n-; I ++ ) {
 21 is              Result [I] = Character.forDigit is (list.remove (K / factorial [n---I. 1]), 10 );
 22 is              K = K% factorial [l--n- I];
 23 is          }
24         return new String(result);
25     }