Permutation Sequence

最新推荐文章于 2019-09-07 16:05:54 发布

原创最新推荐文章于 2019-09-07 16:05:54 发布 · 173 阅读

0 ·

CC 4.0 BY-SA版权

算法专栏收录该内容

636 篇文章

订阅专栏

本文介绍了一种算法，用于找出从1到n的所有唯一排列中的第k个排列序列。通过递归方法遍历所有可能的排列组合，并在过程中计数以确定何时找到目标序列。

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 (ie, for n = 3):

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

Given n and k, return the k^th permutation sequence.

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

void visit(int n, int k, int pos, int &count, bool flag[], int result[])
{
	if (count == k)
	{
		return;
	}

	if (pos == n)
	{
		count++;
		if (count == k)
		{
			for (int i = 0; i < n; i++)
			{
				cout << result[i];
			}
			cout << endl;
		}
		return;
	}

	for (int i = 0; i < n; i++)
	{
		if (flag[i])
		{
			result[pos] = i+1;
			flag[i] = false;
			visit(n, k, pos+1, count, flag, result);
			flag[i] = true;
		}
	}
}

void fun(int n, int k)
{
	bool flag[n];
	memset(flag, true, sizeof(flag));
	int result[n];
	int count = 0;
	visit(n, k, 0, count, flag, result);
}