Sunday, October 4, 2015

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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