60. 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 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.
• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

 

AC code:

class Solution {
public:
    string getPermutation(int n, int k) {
        string s(n, 0);
        iota(s.begin(), s.end(), '1');
        vector<int> fact(n, 1);
        string ans(n, 0);
        for (int i = n-3; i >= 0; --i)
            fact[i] = (n - 1 - i) * fact[i+1];
        k = k - 1;
        for (int i = 0; i < n; ++i) {
            int index = k / fact[i];
            k = k % fact[i];
            ans[i] = s[index];
            s.erase(next(s.begin(), index));
        }
        return ans;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Permutation Sequence.

 

std::iota

template <class ForwardIterator, class T>
  void iota (ForwardIterator first, ForwardIterator last, T val);

Store increasing sequence

Assigns to every element in the range [first,last) successive values of val, as if incremented with ++val after each element is written.

The behavior of this function template is equivalent to:

1 2 3 4 5 6 7 8 9

template <class ForwardIterator, class T> void iota (ForwardIterator first, ForwardIterator last, T val) { while (first!=last) { *first = val; ++first; ++val; } }

Parameters

first, last

Forward iterators to the initial and final positions of the sequence to be written. The range used is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last.

val

Initial value for the accumulator.