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):
"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.
這道題其實(shí)有很強(qiáng)的規(guī)律可循。首先，n個(gè)元素的排列總數(shù)是n!。在下面的分析中，讓k的范圍是0 <= k < n!。（題目代碼實(shí)際上是1<=k<=n!)
可以看到一個(gè)規(guī)律，就是這n！個(gè)排列中，第一位的元素總是(n-1)!一組出現(xiàn)的，也就說如果p = k / (n-1)!，那么排列的最開始一個(gè)元素一定是arr[p]。
這個(gè)規(guī)律可以類推下去，在剩余的n-1個(gè)元素中逐漸挑選出第二個(gè)，第三個(gè)，...，到第n個(gè)元素。程序就結(jié)束。
1 /**
2 * The set [1,2,3,…,n] contains a total of n! unique permutations.
3 *
4 * By listing and labeling all of the permutations in order, We get the
5 * following sequence (ie, for n = 3):
6 *
7 * "123" "132" "213" "231" "312" "321" Given n and k, return the kth permutation
8 * sequence.
9 *
10 * Note: Given n will be between 1 and 9 inclusive.
11 *
12 */
13
14 public class PermutationSequence {
15 public String getPermutation(int n, int k) {
16 char[] arr = new char[n];
17 int pro = 1;
18 for (int i = 0; i < n; ++i) {
19 arr[i] = (char) ('1' + i);
20 pro *= (i + 1);
21 }
22 k = k - 1;
23 k %= pro;
24 pro /= n;
25 for (int i = 0; i < n - 1; ++i) {
26 int selectI = k / pro;
27 k = k % pro;
28 pro /= (n - i - 1);
29 int temp = arr[selectI + i];
30 for (int j = selectI; j > 0; --j) {
31 arr[i + j] = arr[i + j - 1];
32 }
33 arr[i] = (char) temp;
34 }
35 return String.valueOf(arr);
36 }
37 }
38