1 class Solution {
 2 public:
 3     string getPermutation(int n, int k) {
 4         string ans(n, ‘0‘);
 5         vector<bool> flag(n, false);
 6         int count = 1;
 7         for (int i = 1; i < n; i++)
 8             count *= i;
 9         int m;
10         int m_f;
11         int index = 0
 
;
12         k = k - 1;
13         while (index <n-1)
14         {
15             m = k / count;
16             m_f = 0;
17             for (int i = 0; i <= m; i++, m_f++)
18             {
19                 while (flag[m_f])
20                     m_f++;
21             }
22             m_f--;
23             flag[m_f] = true
 
;
24             ans[index] = m_f + 1+ ‘0‘;
25             index++;
26             k = k % count;
27             count = count / (n - index );
28         }
29         int i = 0;
30         while (flag[i])
31             i++;
32         ans[n - 1] = i + 1 + ‘0‘;
33         return ans;
34         
35         
36
 
     }
37 };

總結：幾乎是用找規律算出來的，再次寫，未必能流暢寫下來。

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

LEETCODE60——第K個排列