leetcode Permutation Sequence

Permutation Sequence

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Total Accepted: 40746 Total Submissions: 174549 Difficulty: Medium

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.

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class Solution {public:    string searchNum(vector<bool> &signal, int count, string res){char temp;int i = 1;for (i; i < signal.size() && count >= 0; i++){if (signal[i] == false)count--;}signal[i-1] = true;temp = i-1 + 48;res += temp;return res;}string getPermutation(int n, int k){vector<int> njie;int count = 0;njie.push_back(1);vector<bool> nsignal;nsignal.push_back(false);for (int i = 1; i <= n; i++){njie.push_back(njie[i - 1] * i);nsignal.push_back(false);}string result = "";k--;//此处特别特别重要，因为所有的都是从0开始计算，所以k要减1int q = n - 1;while (q>=0){count = k / njie[q];k = k%njie[q];q--;result = searchNum(nsignal, count, result);}return result;}};