Permutation Sequence 解答

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.

Solution -- Math

We can conclude a pattern by obeservation.

[x₁, x₂, x₃, .., x_n]

If we fix x₁, then number of permutations is (n - 1)!

If we fix x₁ and x₂, then number of permutations is (n - 2)!

Following this thinking way, we can solve this problem by finding x_i iteratively.

We also need a visited array here to note which element has been used.

 public class Solution {
     public String getPermutation(int n, int k) {
         StringBuilder sb = new StringBuilder(n);
         if (n < 1 || n > 9)
             return sb.toString();
         int factorial = 1;
         for (int i = 1; i <= n; i++) {
             factorial *= i;
         }
         if (k > factorial || k < 1)
             return sb.toString();
         boolean[] visited = new boolean[n];

         int num = n;
         int start = 1;
         while (num > 0) {
             factorial /= num;
             start += (k / factorial);
             if (k % factorial == 0)
                 start -= 1;
             int index = 0;
             for (int i = 1; i <= n; i++) {
                 // Find the right index
                 if (!visited[i - 1])
                     index++;
                 if (index == start) {
                     sb.append(i);
                     visited[i - 1] = true;
                     break;
                 }
             }
             k = k - (start - 1) * factorial;
             start = 1;
             num--;
         }
         return sb.toString();
     }
 }