60. Permutation Sequence

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

link: link

https://leetcode.com/problems/permutation-sequence/discuss/22507/%22Explain-like-I'm-five%22-Java-Solution-in-O(n)

Solution from Forum

public String getPermutation(int n, int k) {
    int pos = 0;
    List<Integer> numbers = new ArrayList<>();
    int[] factorial = new int[n+1];
    StringBuilder sb = new StringBuilder();

    // create an array of factorial lookup
    int sum = 1;
    factorial[0] = 1;
    for(int i=1; i<=n; i++){
        sum *= i;
        factorial[i] = sum;
    }
    // factorial[] = {1, 1, 2, 6, 24, ... n!}

    // create a list of numbers to get indices
    for(int i=1; i<=n; i++){
        numbers.add(i);
    }
    // numbers = {1, 2, 3, 4}

    k--;

    for(int i = 1; i <= n; i++){
        int index = k/factorial[n-i];
        sb.append(String.valueOf(numbers.get(index)));
        numbers.remove(index);
        k-=index*factorial[n-i];
    }

    return String.valueOf(sb);
}