lintcode-medium-Permutation Sequence

Given n and k, return the k-th permutation sequence.

Notice

n will be between 1 and 9 inclusive.

Example

For

n = 3

, all permutations are listed as follows:

"123"
"132"
"213"
"231"
"312"
"321"

If

k = 4

, the fourth permutation is

"231"

Challenge

O(n*k) in time complexity is easy, can you do it in O(n^2) or less?

class Solution {
/**
* @param n: n
* @param k: the kth permutation
* @return: return the k-th permutation
*/
public String getPermutation(int n, int k) {
if(n <= 0)
return "";

StringBuilder result = new StringBuilder();
ArrayList<Integer> list = new ArrayList<Integer>();

for(int i = 1; i <= n; i++)
list.add(i);

k--;
n--;

while(n >= 0){
int index = k / factor(n);

result.append(list.remove(index));
k -= index * factor(n);
n--;
}

return result.toString();
}

public int factor(int n){
int result = 1;

for(int i = 1; i <= n; i++)
result *= i;

return result;
}
}