[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 kth permutation sequence.

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

Solution:

class Solution {

public:

    string getPermutation(int n, int k) {

    string ans = "";

    int *fact = new int[n + 1];

    int *data = new int[n + 1];

    fact[0] = 1;

    fact[1] = 1;

    for(int i = 1;i <= n;i++)

    {

        data[i] = i;

        fact[i] = fact[i - 1] * i;

    }



    int tmpK = k;

    for(int i = 1; i <= n;i++)

    {

        int ind = (tmpK - 1) / fact[n - i] + 1;

        tmpK = (tmpK - 1) % fact[n - i] + 1;

        int num_loc = 0;//current number's index in the remaining number list

        for(int j = 1;j <= n;j++)

        {

            if(data[j] != 0)

            {

                num_loc++;

                if(num_loc == ind)

                {

                    //this is the desire number for this location

                    ans += data[j] + '0';

                    data[j] = 0;

                    break;

                }

            }

        }

    }



    return ans;

    }

};