Leetcode NO.60 Permutation Sequence

题目要求如下：

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.

算法的思路如下：

1，定义一个map存储 1到 n这n个值。C++ 的STL map是根据key大小排序的

2，要高清排序规则，就拿题干中的例子举例：

比如第三位213，如果从零开始排位，它就是k = k - 1 =2;

2.1 k / (n-1)! = 2 / 2 = 1, map里面有三个key 1,2,3选择序号是1的即2，然后删除此值, k = k % (n-1)! = 0

2.2 0 / (n-2)! = 0 / 1 = 0, map里面有两个key 1,3选择序号是0的即1，然后删除此值，k = k % (n-2)! = 0

2.3 只剩一个值，就是3.

213为最终解，

代码如下：

class Solution {public:    string getPermutation(int n, int k) {    string retVal = "";        map<int, int> num_map;        map<int, int>::iterator it;        for (int i = 1; i <= n; ++i)        num_map[i] = i;        int tmp_n = n - 1;        k = k - 1;        while (tmp_n >= 0) {        int tmp = k / factorial(tmp_n);        k = k % factorial(tmp_n);        for (it = num_map.begin(); it != num_map.end(); ++it) {        if (tmp == 0){        retVal += to_string(it->first);        num_map.erase(it);        break;        }        else        --tmp;        }        --tmp_n;        }        return retVal;    }private:int factorial(int n) {int result = 1;for (int i = 1; i <= n; ++i)result *= i;return result;}};