leetcode 74: Permutation Sequence

Permutation SequenceMar 28 '12

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 k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

 

 

 

public class Solution {
    public String getPermutation(int n, int k) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(k<0) return new String();
        
        k = k-1;
        
        LinkedList<Character> can = new LinkedList<Character>();
        for(int i=0; i<n;i++) {
            can.add( (char)(i+'1') );
        }
        
        char[] res = new char[n];
        int i=0;
        int p=0;
        
        while(i<n) {
            int f = getFactorial(n-i-1);
            p = k/f;
            res[i++] = can.get(p);
            can.remove(p);
            k = k % f; 
        }
        
        return new String(res);
    }
    
    private int getFactorial(int n) {
        if(n==0) return 1;
        int x = n;
        while(--n>0) x *= n;
        return x;
    }
}