leetcode：Permutation Sequence

最新推荐文章于 2019-08-10 11:01:42 发布

PAUL_shuo

最新推荐文章于 2019-08-10 11:01:42 发布

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本文采用康托编码的方法，解决求解排列序列的问题。通过给出的数学原理和代码实现，详细介绍了如何从给定的排列数k找到对应的排列顺序。适用于对组合数学和算法有兴趣的读者。

Permutation Sequence Total Accepted: 3357 Total Submissions: 16092 My Submissions
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 kth permutation sequence.

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

利用康托编码的思路，假设有n 个不重复的元素，第k 个排列是a1; a2; a3; :::; an，那么a1 是
哪一个位置呢？
代码：

class Solution {
public:
	string getPermutation(int n, int k)
	{
		int ResArray[10];
		int finalArray[10];
		for (size_t i =n; i >=1; i--)
		{
			ResArray[i-1] = (k-1)/numArray[i-1];
			k -= ResArray[i-1] * numArray[i-1];
		}
		for (size_t i = n; i >=2; i--)
		{
			finalArray[i] = sArray[ResArray[i-1] + 1];
			con(ResArray[i-1] + 1);

		}
		finalArray[1] = sArray[1];
		int final = finalArray[n];
		for (int i = n-1; i >=1; i--)
		{
			final = final * 10 + finalArray[i];
		}
		stringstream ss;
		ss << final;
		return ss.str();

		

	}
public:
	int mul(int n)
	{
		int k = 1;
		while (n >= 2)
		{
			k *= n;
			n--;
		}
		return k;
	}
public:
	void con(int n)
	{
		for (size_t i = n; i < 9; i++)
		{
			sArray[i] = sArray[i + 1];
		}
	}
private: int numArray[9];
private: int sArray[10];
public: Solution()
		 {
			 sArray[9] = 9;
			 for (size_t i = 1; i <= 8; i++)
			 {
				 numArray[i] = mul(i);
				 sArray[i] = i;
			 }
		 }
};