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.

Have you met this question in a real interview? Yes
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;
    }
}