The set 

[1,2,3,…,n]

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.

求由n个数字组成的有序的排列中的第k个排列，因为只需返回第k个，所以不用将所有的排列组合都求出来，只求出第k个排列组合即可，重点在于找出排序和k的规律。

参考：喜刷刷

Python：

class Solution(object):
def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
seq, k, fact = "", k - 1, math.factorial(n - 1)
perm = [i for i in xrange(1, n + 1)]
for i in reversed(xrange(n)):
curr = perm[k / fact]
seq += str(curr)
perm.remove(curr)
if i > 0:
k %= fact
fact /= i
return seq　

C++：

class Solution {
public:
string getPermutation(int n, int k) {
string ret;
vector<int> factorial(n,1);
vector<char> num(n,1);

for(int i=1; i<n; i++)
factorial[i] = factorial[i-1]*i;

for(int i=0; i<n; i++)
num[i] = (i+1)+'0';

k--;
for(int i=n; i>=1; i--) {
int j = k/factorial[i-1];
k %= factorial[i-1];
ret.push_back(num[j]);
num.erase(num.begin()+j);
}

return ret;
}
}; 

C++：

class Solution {
public:
string getPermutation(int n, int k) {
string res;
string num = "123456789";
vector<int> f(n, 1);
for (int i = 1; i < n; ++i) f[i] = f[i - 1] * i;
--k;
for (int i = n; i >= 1; --i) {
int j = k / f[i - 1];
k %= f[i - 1];
res.push_back(num[j]);
num.erase(j, 1);
}
return res;
}
};

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[LeetCode] 46. Permutations 全排列

[LeetCode] 47. Permutations II 全排列 II 

[LeetCode] 31. Next Permutation 下一个排列

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