leetcode

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 will be between 1 and 9 inclusive.

Solution:

  public String getPermutation(int n, int k) {    List<Integer> numbers = new ArrayList<>();    int[] factorial = new int[n + 1];    StringBuilder sb = new StringBuilder();    int sum = 1;    factorial[0] = 1;    for (int i = 1; i <= n; i++) {      sum *= i;      factorial[i] = sum;    }    // factorial[] = {1, 1, 2, 6, 24, ... n!}    for (int i = 1; i <= n; i++) {      numbers.add(i);    }    // numbers = {1, 2, 3, 4}    k--;    for (int i = 1; i <= n; i++) {      int index = k / factorial[n - i];      sb.append(numbers.get(index));      numbers.remove(index);      k -= index * factorial[n - i];    }    return String.valueOf(sb);  }