solution.md

60. Permutation Sequence

Difficulty: Medium

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 k^th 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"

Solution

Language: Java

class Solution {
    public String getPermutation(int n, int k) {
        int[] factorial = new int[n + 1];
        List<Integer> leftNumList = new ArrayList<>();
        int sum = 1;
        factorial[0] = 1;
        for (int i = 1; i <= n; i++) {
            leftNumList.add(i);
            sum *= i;
            factorial[i] = sum;
        }
        StringBuilder stringBuilder = new StringBuilder();
        k--; // Amazing k--
        for (int i = 1; i <= n; i++) {
            int index = k / factorial[n - i];
            k = k % factorial[n - i];
            stringBuilder.append(leftNumList.get(index));
            leftNumList.remove(index);
        }
        return stringBuilder.toString();
    }
}