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 for n = 3:

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

Given n and k, return the kth permutation sequence.

Example 1:
Input: n = 3, k = 3
Output: "213"


Example 2:
Input: n = 4, k = 9
Output: "2314"

找出n个数全排列的第k个序列。

解决：n个数全排列，第1~(n-1)!个数，肯定是1打头的。

用一个数组记录一下已经使用过的数字。

class Solution {
public:
    int Allper(int n) {
        int res = 1;
        while (n > 0)
            res *= n--;
        return res;
    }
    string getPermutation(int n, int k) {
        int cnt = 0;
        vector<int> num(n);
        string res;
        int i;
        int total;
        while (res.size() < n) {
            total = Allper(n - res.size() - 1);
            i = 0;
            while (i * total + cnt < k)
                ++i;
            cnt += (i - 1) * total;
            for (int j=0; j<n; ++j) {
                if (num[j] == 0)
                    --i;
                if (i == 0) {
                    num[j] = 1;
                    res += to_string(j + 1);
                    break;
                }
            }
        }
        return res;
    }
};