Buttercola: Zenefits: [OA] Word Rank
给一组String[](容量<10000)，求每一个String(长度<100)在其自己的Permutation Sequence中的序号
Input: // String[]
cab
babc
Output: //返回 int[], index starting from 0
4 //{abc, acb, bac, bca, 【cab】, cba}
3 //{abbc, abcb, acbb, 【babc】,.....}
//注意，String内有重复的Character，但 Permutation Sequence 只保留distinct记录
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