60. Permutation Sequence java solutions

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.

public class Solution {
    public String getPermutation(int n, int k) {
        StringBuilder ans = new StringBuilder();
        StringBuilder tmp = new StringBuilder();
        for(int i = 1; i<=n; i++) tmp.append(i);
        int factor = calculate(n);
        int index = 0;
        for(int i = n; i>=1; i--){
            factor /= i;
            index = (k-1)/factor;
            ans.append(tmp.charAt(index));
            k -= index*factor;
            tmp.deleteCharAt(index);
        }
        return ans.toString();
    }
    
    public int calculate(int n){
        int sum = 1;
        for(;n>=1; n--){
            sum *= n;
        }
        return sum;
    }
}

n个数的permutation总共有n阶乘个，基于这个性质我们可以得到某一位对应的数字是哪一个。思路是这样的，比如当前长度是n，我们知道每个相同的起始元素对应(n-1)!个permutation，

也就是(n-1)!个permutation后会换一个起始元素。因此，只要当前的k进行(n-1)!取余，得到的数字就是当前剩余数组的index，如此就可以得到对应的元素。

https://discuss.leetcode.com/topic/17348/explain-like-i-m-five-java-solution-in-o-n