# 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 (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.

## Solution

1. Brute!
2. combinatorial mathematics.

## Code

``````public void nextPermutation(char[] num) {
int last = num.length - 1;
int i = last;
while (i > 0 && num[i - 1] >= num [i]) --i;
for (int l = i, r = last; l < r; ++l, --r) {
num[l] = (char) (num[l] ^ num[r]);
num[r] = (char) (num[l] ^ num[r]);
num[l] = (char) (num[l] ^ num[r]);
}
if (i == 0) {
return;
}
int j = i;
while (j <= last && num[i-1] >= num[j]) ++j;
num[i-1] = (char) (num[i-1] ^ num[j]);
num[j] = (char) (num[i-1] ^ num[j]);
num[i-1] = (char) (num[i-1] ^ num[j]);
}

public String getPermutation_1(int n, int k) {
char[] num = new char[n];
for (int i = 0; i < n; ++i) num[i] = (char) (i + '1');
System.out.println(String.valueOf(num));
while (--k != 0) {
nextPermutation(num);
}
return String.valueOf(num);
}

public String getPermutation_2(int n, int k) {
StringBuffer sb = new StringBuffer();
StringBuffer res = new StringBuffer();
int total = 1;
for (int i = 1; i <= n; ++i) {
total = total * i;
sb.append(i);
}
k--;
while(n != 0) {
total = total / n;
int idx = k / total;
res.append(sb.charAt(idx));
k = k % total;
sb.deleteCharAt(idx);
n--;
}
return res.toString();
}
``````