Problem

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:

• "123"
• "132"
• "213"
• "231"
• "312"
• "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"

Solution

/**
 * @param {number} n
 * @param {number} k
 * @return {string}
 */
var getPermutation = function(n, k) {
  var str = '';
  var nums = [1, 2, 3, 4, 5, 6, 7, 8, 9];
  var factorial = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]; // n!
  var tmp1 = 0;
  var tmp2 = 0;

  k--;

  for (var j = n; j >= 1; j--) {
    tmp1 = factorial[j - 1];
    tmp2 = Math.floor(k / tmp1);

    k %= tmp1;
    str += nums[tmp2];

    nums.splice(tmp2, 1);
  }

  return str;
};

Explain:

用回溯的方法会超时， 需要用数学规律解的.就是依次找出放在第 index 位的数字。

k-- 这里比较难理解，是用来纠正 k 正好整除于某个阶乘数的特殊情况，即 k % factorial[i] === 0 时。

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).