public?static?void?main(String[]?args)?{
????System.out.println(lexicographicPermu(10,?1000000L));
    }

    private?static?String?lexicographicPermu(int?numberOfDigit,?long?target)?{
????int?copyOfNumerOfDigit?=?numberOfDigit;
????long?copyOfTarget?=?target;

????//?list?里面存放的是所有的可以选择的情况，并从大到小排列
????//?这里是?0，1，2，3，4，5，6，7，8，9
????List<Integer>?candidate?=?new?ArrayList<>();
????for?(int?i?=?0;?i?<?copyOfNumerOfDigit;?++i)?{
????????candidate.add(i);
????}

????String?result?=?"";

????//?只要没有选完，就继续选择
????while?(copyOfNumerOfDigit?!=?0)?{
????????long?totalPermutation?=?1;
????????int?n?=?copyOfNumerOfDigit;

????????//?计算?n!
????????//?例如，选择第一个数，那?totalPermutation?就是?10!
????????while?(n?!=?1)?{
????????????totalPermutation?*=?n;
????????????n--;
????????}

????????//?计算选择当前情况下的第几个数
????????//?这里?copyOfTarget?-?1?是为了防止整除的情况
????????//?举?0，1，2?这个例子，当我们要选择排在第?2?的那个数，也就是?021
????????//?选第一个数的时候，?totalPermutation?=?3?*?2?*?1?=?6,?copyOfNumberOfDigit?=?3
????????//?2?/?(6?/?3)?=?1，显然不符合要求，(2?-?1)?/?(6?/?3)?=?0?才是选中了?0
????????int?selection?=?(int)((copyOfTarget?-?1)?/?(totalPermutation?/?copyOfNumerOfDigit));
????????result?+=?candidate.remove(selection);

????????//?更新下一次要选择第几个数
????????copyOfTarget?-=?(totalPermutation?/?copyOfNumerOfDigit)?*?selection;

????????//?选了一个数，可选项相应减少一
????????copyOfNumerOfDigit--;
????????copyOfTarget?=?copyOfTarget?<=?0???0?:?copyOfTarget;
????}

????return?result;
}