60. Permutation Sequence

description:

找到全排列中的第k個序列
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:

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the kth permutation sequence.
Note:

Given n will be between 1 and 9 inclusive.
Given k will be between 1 and n! inclusive.

Example:

Example 1:

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

Example 2:

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

answer:

class Solution {
public:
    string getPermutation(int n, int k) {
        string res;
        string num = "123456789";
        vector<int> f(n, 1);
        for (int i = 1; i < n; ++i) f[i] = f[i - 1]*i;
        --k;
        for (int i = n; i >= 1; --i) {
            int j = k / f[i - 1];
            k %= f[i - 1];
            res.push_back(num[j]);
            num.erase(j, 1);
        }
        return res;
    }
};

relative point get√：

https://blog.csdn.net/u010472607/article/details/80431604 【erase】

hint :

http://www.javashuo.com/article/p-fbxopymn-gk.html