Permutation Sequence 解答

Question

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.

Solution -- Math

We can conclude a pattern by obeservation.

[x₁, x₂, x₃, .., x_n]

If we fix x₁, then number of permutations is (n - 1)!

If we fix x₁ and x₂, then number of permutations is (n - 2)!

Following this thinking way, we can solve this problem by finding x_i iteratively.

We also need a visited array here to note which element has been used.

public class Solution {
    public String getPermutation(int n, int k) {
        StringBuilder sb = new StringBuilder(n);
        if (n < 1 || n > 9)
            return sb.toString();
        int factorial = 1;
        for (int i = 1; i <= n; i++) {
            factorial *= i;
        }
        if (k > factorial || k < 1)
            return sb.toString();
        boolean[] visited = new boolean[n];

        int num = n;
        int start = 1;
        while (num > 0) {
            factorial /= num;
            start += (k / factorial);
            if (k % factorial == 0)
                start -= 1;
            int index = 0;
            for (int i = 1; i <= n; i++) {
                // Find the right index
                if (!visited[i - 1])
                    index++;
                if (index == start) {
                    sb.append(i);
                    visited[i - 1] = true;
                    break;
                }
            }
            k = k - (start - 1) * factorial;
            start = 1;
            num--;
        }        
        return sb.toString();
    }
}