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.

   思路，使用字典序法，与http://blog.csdn.net/mlweixiao/article/details/38897499
   

   public class Solution {
   	
   	private void nextPermutation(int[] num) {
   		int i;
   		int cur = -1;
   		int temp;
   		// find the last increase sequence
   		for (i = num.length - 1; i >= 1; i--) {
   			if (num[i] > num[i - 1]) {
   				cur = i - 1;
   				break;
   			}
   		}
   
   		// if the increase sequence exists,
   		// swap the cur and the last one(bigger than it)
   		if (cur != -1) {
   			for (i = num.length - 1; i > cur; i--) {
   				if (num[i] > num[cur]) {
   					temp = num[cur];
   					num[cur] = num[i];
   					num[i] = temp;
   					break;
   				}
   			}
   		}
   
   		for (i = cur + 1; 2 * i <= cur + num.length - 1; i++) {
   			temp = num[i];
   			num[i] = num[num.length - i + cur];
   			num[num.length - i + cur] = temp;
   		}
   	}
   	
       public String getPermutation(int n, int k) {
           int [] temp=new int[n];
           StringBuffer s=new StringBuffer("");
           
           for(int i=0;i