[Leetcode]#60 Permutation Sequence

最新推荐文章于 2025-08-09 17:01:50 发布

原创最新推荐文章于 2025-08-09 17:01:50 发布 · 269 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #string

string 专栏收录该内容

11 篇文章

订阅专栏

本文提供了一段使用C++实现的代码，用于计算给定n和k时的排列序列中的第k个元素。代码通过递归方式计算阶乘并逐次确定序列中的数字。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

//#60 Permutation Sequence
//4ms 28.74%
class Solution {
public:
    string getPermutation(int n, int k) 
    {
        string result;
        if(k > getFactorial(n)) return result;
        vector<int> nums(n, 0);
        for(int i=0; i<n; i++)
        {
            nums[i] = i+1;
        }

        for(int i=0; i<n; i++)
        //determine the corresponding item one by one
        {
            //cout << "Current session is " << n-i << endl;
            //cout << "k = " << k << endl;
            int insert_num = (k-1) / getFactorial(n-i-1);
            //cout << "insert_num = " << insert_num << endl;
            if(insert_num > 0)
            {
                result.push_back( nums[insert_num] + '0' );
                //cout << "inserting " << nums[insert_num] << " into string\n";
                nums.erase(nums.begin()+insert_num);
                k = k - insert_num * getFactorial(n-i-1);
            }
            else
            {
                result.push_back( nums[0] + '0' );
                nums.erase(nums.begin());
            }
        }


        return result;
    }

    int getFactorial(int n)
    {
        if(n==1 || n==0) return 1;
        else return n*getFactorial(n-1);
    }
};