60. 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 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"

难度：medium

题目：集合[1, 2, 3, .... n]包含n!个不同的排列。有序的列出所有排列
给定n和k, 返回第k个序列。

思路：结合next_permutation。此方法非简单方法。

Runtime: 31 ms, faster than 16.21% of Java online submissions for Permutation Sequence.
Memory Usage: 37.6 MB, less than 6.50% of Java online submissions for Permutation Sequence.

class Solution {
    public String getPermutation(int n, int k) {
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = i + 1;
        }
        for (int i = 0; i < k - 1; i++) {
            nextPermutation(nums);
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            sb.append(nums[i]);
        }
        
        return sb.toString();
    }
    
    public void nextPermutation(int[] nums) {
        if (null == nums || nums.length <= 1) {
            return;
        }
        int right = nums.length - 1, j = right;
        for (; j > 0; j--){
            if (nums[j - 1] < nums[j]) break;
        }
        for (int k = right; k >= 0; k--) {
            if (j > 0 && nums[k] > nums[j - 1]) {
                swap(nums, j - 1, k);
                break;
            }
        }
        for (; j < right; j++, right--) {
            swap(nums, j, right);
        }
    }
    
    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}