Lintcode: Permutation Index II

Given a permutation which may contain repeated numbers, find its index in all the permutations of these numbers, which are ordered in lexicographical order. The index begins at 1.

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Example
Given the permutation [1, 4, 2, 2], return 3.

这里需要考虑重复元素，有无重复元素最大的区别在于原来的1!, 2!, 3!...等需要除以重复元素个数的阶乘。记录重复元素个数同样需要动态更新，引入哈希表这个万能的工具较为方便。

按照从数字低位到高位进行计算

 public class Solution {
     /**
      * @param A an integer array
      * @return a long integer
      */
     public long permutationIndexII(int[] A) {
         // Write your code here
         if (A==null || A.length==0) return new Long(0);
         int len = A.length;
         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
         long index = 0, fact = 1, mulFact = 1;
         for (int i=len-1; i>=0; i--) {
             if (!map.containsKey(A[i])) {
                 map.put(A[i], 1);
             }
             else {
                 map.put(A[i], map.get(A[i])+1);
                 mulFact *= map.get(A[i]);
             }
             int count = 0;
             for (int j=i+1; j<len; j++) {
                 if (A[i] > A[j]) count++;
             }
             index += count*fact/mulFact;
             fact *= (len-i);
         }
         index = index + 1;
         return index;
     }
 }