LintCode_198 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.

Example

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

思路是对于每一个字符，找到所有比他小的字符的集合， 然后对于集合中的每个字符，先把他去掉，查看剩下下有多少种作何方式（用排列组合去重）然后再把那个字符添回来；

每个字符完成后，删掉后进行下一个字符的运算：

c++

class Solution {
public:
    /**
     * @param A an integer array
     * @return a long integer
     */
    long long permutationIndexII(vector& A) {
        // Write your code here
        // Write your code here
        vector B = A;
        const int len = A.size();
        vector dp = {0};
        long long res = 1;
        for (long long i = 1; i< len; i++) {
            if (i == 1)
                dp.push_back(1);
            else
                dp.push_back(dp[i - 1] * i);
        }
        sort(B.begin(), B.end());
        for (int i = 0; i < A.size(); i++) {
            auto itr = lower_bound(B.begin(), B.end(), A[i]);
            int ind = itr - B.begin();
            int fake_head = INT_MIN;
            for (int j = 0; j < ind; j++) { 
                if (B[j] != fake_head) {
                   int k = B[j];
                   B.erase(B.begin() + j);
                   res+=get_permutation_num(B, dp);
                   B.insert(B.begin() + j, k);
                   fake_head = B[j];
                }
            }
            //cout<& vec, vector& dp) {
        sort(vec.begin(), vec.end());
        vector dup_num;
        long long res = 1;
        int acl_num = 1;
        int dup = vec[0];
        for (int i = 1; i < vec.size(); i++) {
            if (vec[i] != dup) {
                dup_num.push_back(acl_num);
                dup = vec[i];
                acl_num = 1;
            } else {
                acl_num++;
            }
            res*=((long long)i + 1); 
        }
        dup_num.push_back(acl_num);
        for (int i = 0; i < dup_num.size(); i++) {
                res/=dp[dup_num[i]];
        }
        return res;
    }
};