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 k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

string getPermutation2(int n, int k) 
{
	string ans;
	ans.resize(n);
	bool used[10];
	memset(used, 0, sizeof(used));
	// 用0到n-1代表（1，n）
	for (int i = n; i <= 9; ++i)
	{
		used[i] = true;
	}
	int count = 1;
	for (int i = 1; i <= n - 1; ++i)
	{
		count *= i;
	}
	int num = 0;
	--k;
	while (k)
	{
		int index = k / count;
		int visited = 0;
		for (int i = 0; i < n; ++i)
		{
			if (!used[i])
			{
				if (visited == index)
				{
					ans[num++] = i + '0' + 1;
					used[i] = true;
					break;
				}
				else
				{
					++visited;
				}
			}
		}
		k = k % count;
		if (k == 0)
			break;

		count = count / (n - num);
	}
	if (num < n)
	{
		for (int i = 0; i < n; ++i)
		{
			if (!used[i])
			{
				ans[num++] = i + '0' + 1;
			}
		}
	}
    return ans;
}

void nextPermutation(vector &num) 
{
	int size = num.size();
	if (size == 0)
		return;
	int i = size - 2;
	for (; i >= 0; --i)
	{
		if (num[i] < num[i+1])
			break;
	}
	if (i == -1)
	{
		for (int i = 0; i < (size >> 1); ++i)
		{
			swap(num[i], num[size - 1 - i]);
		}
		return;
	}
	int j = size - 1;
	int minNum = num[i + 1];
	int index = i + 1;
	for (; j > i; --j)
	{
		if (num[j] > num[i])
		{
			if (num[j] < minNum)
			{
				minNum = num[j];
				index = j;
			}
		}
	}
	swap(num[i], num[index]);
	sort(num.begin() + i + 1, num.end());
}