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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):
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
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void permute(int n, int k, int idx, int &count, string &solution, string &res, vector<bool> &isVisited)
{
if(idx == n)
{
count++;
if(count == k)
res.assign(solution);
return;
}
for(int i = 0; i < n; i++)
{
if(!isVisited[i])
{
isVisited[i] = true;
solution.append(1, i + '1');
permute(n, k, idx + 1, count, solution, res, isVisited);
isVisited[i] = false;
solution.assign(solution.substr(0, solution.size() - 1));
}
}
}
string getPermutation(int n, int k)
{
if(n == 0)
return "0";
string solution, res;
vector<bool> isVisited(n, false);
int count = 0;
permute(n, k, 0, count, solution, res, isVisited);
return res;
}