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docs/Leetcode_Solutions/C++/0060._Permutation_Sequence.md

+# 60. Permutation Sequence
+
+**<font color=red>难度:Medium<font>**
+
+## 刷题内容
+> 原题连接
+
+* https://leetcode.com/problems/permutation-sequence/
+
+> 内容描述
+
+```
+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"
+```
+
+> 思路1
+******- 时间复杂度: O(n^2)******- 空间复杂度: O(n)******
+
+先将所有数按顺序存在有个字符串中，然后循环n次，每次找到正确的数，由于earse时间复杂度为n，所以总的时间复杂度为n^2
+
+```cpp
+class Solution {
+public:
+    string getPermutation(int n, int k){
+        int arr[n + 1];
+        string ans,org;
+        for(int i = 1;i <= n;++i)
+        {
+            org.push_back(i + '0');
+            i == 1 ? arr[i] = 1 : arr[i] = arr[i - 1] * i;
+        }
+        for(int i = n - 1;i > 0;--i)
+        {
+            int t = k / arr[i];
+            cout<< t<< endl;
+            k = (k) % arr[i];
+            if(!k)
+            {
+                k = arr[i];
+                t--;
+            }
+            ans.push_back(org[t]);
+            org.erase(org.begin() + t);
+            arr[i] = 0;
+        }
+        ans.push_back(org[0]);
+        return ans;
+    }
+};
+```
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