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.

Note:

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

Example 1:

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

Example 2:

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

找出第k排列数，

思路一：将所有排列数都计算出来放到一个数组中，寻找第k个。该方法超时

思路二：利用next_perm来计算下一个排列数，寻找k次即可。

参考代码如下：

注释里为STL代码

class Solution {
public:
    string getPermutation(int n, int k) {
        string str;
        for (int i = 1; i <= n; ++i)
            str += '0' + i;
        while (--k)
        {
            // next_permutation(str.begin(), str.end());
            nextPermutation(str);
        }
        return str;
    }
    
    void nextPermutation(string& nums) {
        if (nums.empty())
            return;
        int n = nums.size();
        int i = 0, j = n-1, k = n-1;
        if (i == j || i > j)
            return;
        while (j > 0)
        {
            if (nums[j] > nums[j-1])
            {
                i = j - 1;
                break;
            }
            j--;
        }
        while (k > i)
        {
            if (nums[k] > nums[i])
            {
                swap(nums[k], nums[i]);
                break;
            }
            k--;
        }
        reverse(nums.begin() + j, nums.end());
    }
};