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.

class Solution {
public:
    //没看懂 下次
    //(k-1)/(i-1)!  (k-1)/(i-1)! 来计算第i位应该用什么数字
    //n=3 k=5 4/2=2  3   2*(3-1)!=4 5-4=1  
    string getPermutation(int n, int k) {
          
    string result;
    vector<int> factorials;
    vector<int> elements;
 
    for(int i = 0; i < n; i++) {
         if(!i) {
             factorials.push_back(1);
         }
         else {
             factorials.push_back(i * factorials[i - 1]);
         }
         elements.push_back(i + 1);
     }
 
    int position = 0;
     for(int i = n - 1; i > -1; i--) {
         position = (k - 1) / factorials[i];
         k = k - position * factorials[i];
         result.push_back('0' + elements[position]);
         elements.erase(elements.begin() + position);
     } 
     return result;
}
    
};