[Leetcode] 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 will be between 1 and 9 inclusive.

最开始是暴力算，但是9！实在是太大了，显然超时，后来网上看到大牛们的算法，从前后向后找，已知对于每一次第n位的数字，都有(n-1)!个排列，所以可以根据这个规律直接找到答案。

 1 class Solution {  

 2 public:  

 3     string getPermutation(int n, int k) {

 4         string str;

 5         vector<int> factorial(n + 1, 1);

 6         for (int i = 1; i <= n; ++i) {

 7             str += i + '0';

 8             factorial[i] = factorial[i-1] * i;

 9         }

10         string perm;

11         --k;    // convert to 0-based index

12         for (int i = n - 1; i >= 0; --i) {

13             int quotient = k / factorial[i];

14             perm += str[quotient];

15             str.erase(quotient, 1);

16             k %= factorial[i];

17         }

18         return perm;

19     }

20 }; 

超时的算法：

 1 class Solution {

 2 public:

 3     string getPermutation(int n, int k) {

 4         string str;

 5         for (int i = 1; i <= n; ++i) {

 6             str.push_back('0' + i);

 7         } 

 8         if (n < 2) return str;

 9         

10         int a, b;

11         while (k--) {

12             for (a = n - 2; a >= 0; --a) 

13                 if (str[a] < str[a+1]) 

14                     break;

15             

16             for (b = n - 1; b > a; --b) 

17                 if (str[b] > str[a])

18                     break;

19             

20             swap(str[a], str[b]);

21             reverse(str.begin() + a + 1, str.end());

22         }

23         return str;

24     }

25 };