60. 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 k^th permutation sequence.

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

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分析：用回溯法似乎太慢，有人用了一种数学的解法。看不懂

class Solution {
public:
    string getPermutation(int n, int k) {
        int Ptable[10]{1};
        for(int t=1;t<10;++t)
        Ptable[t]= t*Ptable[t-1];
        vector<string> v1;
        for(int i=1;i<10;++i)
        {
            v1.push_back(to_string(i));
        }
        string result="";
        while(n>0)
        {
            int temp=(k-1)/Ptable[n-1];
            result+=v1[temp];
            v1.erase(v1.begin()+temp);
            k=k-temp*Ptable[n-1];
            --n;
        }
        return result;
    }
};