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;
    }
}