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: Math

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  string getPermutation(int n, int k) {
    vector<int> num;
    vector<int> fact(10, 1);
    for (int i = 1; i <= 9; i++) {
      num.push_back(i);
      fact[i] = fact[i - 1] * i;
    }

    string s;
    k--;
    while (n--) {
      int d = k / fact[n];
      k %= fact[n];
      s += ('0' + num[d]);
      for (int i = d + 1; i <= 9; i++)
        num[i - 1] = num[i];
    }
    return s;
  }
};

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