Permutation Index II

LintCode

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.

Example

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

解题思路

题 Permutation Index 的扩展，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，颇有点高中排列组合题的味道。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。

 public long permutationIndexII(int[] A) {
        if (A == null || A.length == 0) return 0L;

        Map<Integer, Integer> hashmap = new HashMap<Integer, Integer>();
        long index = 1, fact = 1, multiFact = 1;
        for (int i = A.length - 1; i >= 0; i--) {
            // collect its repeated times and update multiFact
            if (hashmap.containsKey(A[i])) {
                hashmap.put(A[i], hashmap.get(A[i]) + 1);
                multiFact *= hashmap.get(A[i]);
            } else {
                hashmap.put(A[i], 1);
            }
            // get rank every turns
            int rank = 0;
            for (int j = i + 1; j < A.length; j++) {
                if (A[i] > A[j]) rank++;
            }
            // do divide by multiFact
            index += rank * fact / multiFact;
            fact *= (A.length - i);
        }

        return index;
}

References

1 Permutation Index II