"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 would be between 1 and 9 inclusive.

My method determines the number of digits from high to low.

Illustrated by the legend N=3:

Build Array v={1,2,3}

Determine the highest bit:

Ind= (k-1)/2

Note: Denominator 2 refers to the number of permutations that each highest bit lasts. Since there are n-1=2 in addition to the highest bit, there are 2 of them.

The IND refers to the range of the highest position that the K-permutation is asked to fall within.

k==1,2, ind==0, the highest bit is v[ind]==1

k==3,4, Ind==1, the highest bit is v[ind]==2

k==5,6, ind==2, the highest bit is v[ind]==3

The rest of the digits are determined on a per-bit basis. Note that K's take-up update.

classSolution { Public:    stringGetpermutation (intNintk) {vector<int> V (N,0);  for(inti =0; I < n; i + +) V[i]= i+1; stringresult ="";  while(n1)        {            intdivisor = FAC (n1); intIND = (K-1)/Divisor; //itoa            stringDigit; Chartemp[ +]; sprintf (temp,"%d", V[ind]); Digit=temp; Result+=Digit;  for(inti = ind+1; I < v.size (); i + +) V[i-1] =V[i];            V.pop_back (); K%=Divisor; if(k = =0) K=Divisor; N--; }        //itoa        stringDigit; Chartemp[ +]; sprintf (temp,"%d", v[0]); Digit=temp; Result+=Digit; returnresult; }    intFacintN) {if(n = =0)            return 1; intresult =1;  for(inti =1; I <= N; i + +) Result*=i; returnresult; }};

"Leetcode" permutation Sequence