LEETCODE60——第K個排列

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 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