[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.

class Solution {
public:
    string getPermutation(int n, int k) {
        string res;
        func(res,n,k);
        return res;
    }
private:
    void func(string &res,int n,int k){ //n表示几位数
        if(n==0)
            return;
        int weight = f(n-1);
        char cur = '1';
       
        int num = (k-1)/weight + 1;
        while( find(res.begin(),res.end(),cur)==res.end() || num!=0){
             if(find(res.begin(),res.end(),cur)==res.end())
               num--;
              if( num != 0) 
                cur += 1;
              else
                  break;
        }
         k %= weight;
        if(k==0)
            k = weight; 
        res.push_back(cur);            
        func(res,n-1,k);
    }

    int f(int n){
       
       if(n == 1 || n==0)
           return 1;
       else
          return n*f(n-1);
    
    }
};
原文地址:https://www.cnblogs.com/Xylophone/p/3920657.html