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:
"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.
- 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"
Solution: Math
Time complexity: O(n)
Space complexity: O(n)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
// Author: Huahua class Solution { public: string getPermutation(int n, int k) { vector<int> num; vector<int> fact(10, 1); for (int i = 1; i <= 9; i++) { num.push_back(i); fact[i] = fact[i - 1] * i; } string s; k--; while (n--) { int d = k / fact[n]; k %= fact[n]; s += ('0' + num[d]); for (int i = d + 1; i <= 9; i++) num[i - 1] = num[i]; } return s; } }; |
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