【Leetcode】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 {
public:
    string getPermutation(int n, int k) {
        string result;
        bool used[10];
        memset(used, false, 10);
        int base[10];
        base[1] = 1;
        for (int i = 2; i <= n; ++i) {
            base[i] = base[i - 1] * i;
        }
        --k;
        for (int i = 0; i < n; ++i) {
            int p = k / base[n - 1 - i];
            for (int j = 1; j <= n; ++j) {
                if (!used[j] && p-- == 0) {
                    result.push_back('0' + j);
                    used[j] = true;
                    break;
                }
            }
            k %= base[n - 1 - i];
        }
        return result;
    }
};

n个数组的全排列中，以1开头的个数(n-1)!个，同样以2..n开头的个数有(n-1)!个，这样根据(k-1)/[(n-1)!]可知第一个数是多少，依次类推...