【leetcode】 Permutation Sequence (middle)

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.

 

思路：给定序号找排列的字符串，肯定不用一个一个求，根据序号来判断每一位上的数字。用一个向量存储 0~n-1的阶乘，用另一个向量vec从小到大存1~n数字， 求第k位的话，我们用k-1(转为从0开始)， 除以（n-1)!  其整数部分就是该位数字在vec的序号。之后在vec中删掉该数字，k2 %= (n-1)! 以此类推

class Solution {

public:

    string getPermutation(int n, int k) {

        vector<int> factorial(n, 1);

        vector<int> vec(n, 1);

        string ans;

        for(int i = 1; i < n; i++)

        {

            factorial[i] = factorial[i - 1] * i;

            vec[i] = i + 1;

        }

        if(k > factorial[n - 1] * n)

            return ans;



        int k2 = k - 1;

        for(int i = n - 1; i >= 0; i--)

        {

            int cur = k2 / factorial[i];

            char c[2];

            c[0] = '0' + vec[cur];

            c[1] = '\0';

            ans.append(c);

            vec.erase(vec.begin() + cur);

            k2 = k2 % factorial[i];

        }

        return ans;

    }

};