[LeetCode]Permutation Sequence

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.

 

这个字符串的规律就是由小到大的数字序列，第k个字符串就是第k大的数字序列字符串，那么暴力遍历是可行的，但是太慢。

还有个方法就是康托展开，时间复杂度O(n)，空间复杂度O(1)。

class Solution {
public:
    string getPermutation(int n, int k) {
        vector<int> numbers;
        for(int i=1;i<=n;i++)
        {
            numbers.push_back(i);
        }
        vector<int> result;
        int b=k-1;
        for(int i=n-1;i>=0;i--)
        {
            int a=b/(jiecheng(i));
            result.push_back(numbers[a]);
            numbers.erase(numbers.begin()+a);
            if(i!=0)
            {
               b=b%(jiecheng(i));
            }
        }
        string result_str;
        for(int i=0;i<result.size();i++)
        {
            char ch=result[i]+48;
            result_str+=ch;
        }
        return result_str;
    }
    int jiecheng(int n)
    {
        int result=1;
        for(int i=1;i<=n;i++)
        {
            result*=i;
        }
        return result;
    }
};