#198 Permutation Index II

题目描述：

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.

题目思路：

这题和#197不同点在于array中可能有重复元素。高中数学概率告诉我们，有重复元素的情况下，重复k个就要从排列组合总数中除以k！。这里我用一个map来记录每个元素和它出现的个数，如果map中某个数对应的个数>1，就在所求得的base*order中除以个数的阶乘（对于每个重复元素都要这么操作）。

Mycode（AC = 16ms）：

class Solution {public:    /**     * @param A an integer array     * @return a long integer     */    long long permutationIndexII(vector<int>& A) {        // Write your code here        return permutationIndex(A, 0) + 1;    }        long long permutationIndex(vector<int>& A, int idx) {        if (idx >= A.size()) {            return 0;        }                map<int, int> helper;        helper[A[idx]] = 1;                // find A[idx] is x-th smallest in the index idx~end        long long order = 0;        for (int i = idx + 1; i < A.size(); i++) {            if (helper.find(A[i]) != helper.end()) {                helper[A[i]]++;            }            else {                helper[A[i]] = 1;            }                        if (A[i] < A[idx]) {                order++;            }        }                // find the number of permutation for A[idx + 1..end]        long long base = 1;        for (int i = 1; i < A.size() - idx; i++) {            base *= (long long)i;        }                // if there is repeat numbers, then reduce the order by factor        int factor = 1;        for (auto it = helper.begin(); it != helper.end(); it++) {            if (it->second > 1) {                factor *= factorial(it->second);            }        }                return base * order/factor + permutationIndex(A, idx + 1);    }        long long factorial(int num) {        long long f = 1;        for (int i = 1; i <= num; i++) {            f *= (long long)i;        }        return f;    }};