/*
以‘‘开头的的排列有n!个
以1开头的排列有(n-1)!个,若k<(n-1)!,则排列以1开头，若(n-1)!<k<2*(n-1)!，则排列以2开头....
以此类推
....
....
*/
class Solution{
    public String getPermutation(int n, int k) {
        int[] factorial = new int[]{1,1,2,6,24,120,720,5040,40320};//0-8的阶乘
        StringBuilder res = new StringBuilder();
        boolean[] used = new boolean[n+1];//表示元素用过没有
        for(int i=n-1;i>0;i--){
            int a = k/factorial[i];
            int r = k%factorial[i];
            if(r != 0){
                a++;
                k = r;
            }else{
                k = factorial[i];
            }
            for(int j=1;j<=n;j++){
                if(!used[j])a--;
                if(a==0){
                    res.append(j);
                    used[j] = true;
                    break;
                }
            }
            
        }
        for(int i=1;i<=n;i++)if(!used[i])res.append(i);
        return res.toString();
    }
}
/*超时的递归。。
class Solution {
    Set<Integer> res;
    List<Integer> result;
    int k, n, flag;
    public void dfs(List<Integer> output){
        if(flag!=0)return;
        if(n == res.size()){
            //System.out.println("set:"+output+"idx:"+idx);
            if(--k==0){
                result = new ArrayList<Integer>(output);
                flag = 1;
            }
            return;
        }
        for(int i=1;i<=n;i++){
            if(!res.add(i))continue;
            //System.out.println("set:"+output);
            output.add(i);
            dfs(output);
            res.remove(i);
            output.remove(output.size()-1);
        }
    }

    public String getPermutation(int n, int k) {
        this.k = k;
        this.n = n;
        flag = 0;
        res = new HashSet<Integer>();
        result = new ArrayList<Integer>();
        dfs(new ArrayList<Integer>());
        StringBuilder sb = new StringBuilder();
        for(int i:result)sb.append(i);
        return sb.toString();
    }
}
*/