[LeetCode#60]Permutation Sequence

Problem:

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 kth permutation sequence.

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

Analysis:

Once you understand the mathematical principle behind this problem, the idea could be very easy.
To successfully implement out the solution, you must equip good programming skills.
Basic idea:
According to the example, the small number should be placed before large number at first. which means:
For num1 (num2 ... num9), 
There should be permutations among range (num2 ... num9), before we change num1. 
Apparently there are 8! permutations in the range(num2 ... num9). This is the same for every num in the combination. 

By taking advantage of this truth, we could calcualte the index for each position. Suppose k start from 0. 
index[0] = k/(n - 1)!
index[1] = k'/(n - 2)! Note: k' = k % (n - 1)

Skill
1. To use divide and mod operation, we make k start from 0. 
Benefit: the calculated index could be directly used for locating numbers. 
k--;

2. Use ArrayList for recording the available nums. (This method is very elegant! By taking the self adjustment property of the ArrayList, we no longer need to record which num was already used or not.) 
for (int i = 1; i <= n; i++) {
    num.add(i);
}
...
while (round >= 0) {
    int index = k / factorial;
    k %= factorial;
    ret.append(num.get(index));
    num.remove(index);
    if (round > 0)
        factorial /= round;
    round --; 
}

Note: 
index = k / factorial;
The index must in the range of [0, round] when enter the while. 
Cause:
k %= factorial; <factorial is (round+1)>
the k is in the range of [(round-1)!, (round)!]

3. Get initial factorial, adjust it along the loop. 
a. get (n-1)!
for (int i = 2; i <= n - 1; i++) {
    factorial *= i;
}
b. adjust it for next round. 
if (round > 0)
    factorial /= round;
round --; 
c. get current index, and prepare k for getting next index
int index = k / factorial;
k %= factorial;

This problem is not hard, but the programming skills it requires are not easy. 

Time complexity:
Apparently, we need to computate the index for n factorial from (n to 1). For each computation, there is a append and remove operation. the remove operation would take O(n), thus the total time complexity is O(n^2).

Solution:

public class Solution {
    public String getPermutation(int n, int k) {
        if (n <= 0 || n > 9)
            return "";
        k--;
        StringBuffer ret = new StringBuffer();
        ArrayList<Integer> num = new ArrayList<Integer> ();
        int factorial = 1; 
        for (int i = 1; i <= n; i++) {
            num.add(i);
        }
        for (int i = 2; i <= n - 1; i++) {
            factorial *= i;
        }
        int round = n - 1;
        while (round >= 0) {
            int index = k / factorial;
            k %= factorial;
            ret.append(num.get(index));
            num.remove(index);
            //avoid the case at last round
            if (round > 0)
                factorial /= round;
            round --; 
        }
        return ret.toString();
    }
}