[LeetCode]60. Permutation Sequence求全排列第k个

/*
    n个数有n!个排列，第k个排列，是以第（k-1）/(n-1)!个数开头的集合中第（k-1）%(n-1)!个数
     */
    public String getPermutation(int n, int k) {
        k--;
        List<Integer> list = new ArrayList<>();
        StringBuilder res = new StringBuilder();
        int count =1;
        //以每个数字开头的集合有多少中排列
        for (int i = 2; i <= n -1; i++) {
            count*=i;
        }
        //记录哪些数字还没用
        for (int i = 1; i <=n ; i++) {
            list.add(i);
        }
        //回合数,也就是小集合的size
        int round = n-1;
        while (round>=0)
        {
            int num = list.get(k/count);
            res.append(num);
            list.remove(k/count);
            if (round>0)
            {
                k = k%count;
                count/=round;
            }
            round--;
        }
        return res.toString();
    }