leetcode60 Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

class Solution {//找规律，统计，判断每一位的字符
public:
    string getPermutation(int n, int k) {
        string ans;
        vector<char> list;
        for(int i=1;i<=n;i++)
            list.push_back(i+'0');

        while(k)
        {
            if(n==1)
            {
                ans+=list[0];
                break;
            }

            int j=getj(n-1);
            int t=k/j;
            int k2=k-t*j;
            t=(k2==0?t:t+1);

            ans+=list[t-1];
            list.erase(list.begin()+(t-1));

            n--;
            if(k2==0)
                k=j;
            else
                k=k2;

        }

        return ans;
    }

    int getj(int n)
    {
        int ans=1;
        for(int i=1;i<=n;i++)
            ans*=i;
        return ans;
    }
};