Create 254._Factor_Combinations.md

docs/Leetcode_Solutions/254._Factor_Combinations.md

+# 254. Factor Combinations
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/factor-combinations/description/
+
+> 内容描述
+
+```
+Numbers can be regarded as product of its factors. For example,
+
+8 = 2 x 2 x 2;
+  = 2 x 4.
+Write a function that takes an integer n and return all possible combinations of its factors.
+
+Note:
+
+You may assume that n is always positive.
+Factors should be greater than 1 and less than n.
+Example 1:
+
+Input: 1
+Output: []
+Example 2:
+
+Input: 37
+Output:[]
+Example 3:
+
+Input: 12
+Output:
+[
+  [2, 6],
+  [2, 2, 3],
+  [3, 4]
+]
+Example 4:
+
+Input: 32
+Output:
+[
+  [2, 16],
+  [2, 2, 8],
+  [2, 2, 2, 4],
+  [2, 2, 2, 2, 2],
+  [2, 4, 4],
+  [4, 8]
+]
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+
+
+就每次遍历到sqrt(n)就够了，再往后面过去其实重复了, beats 62.6%
+
+```python
+class Solution(object):
+    def getFactors(self, n):
+        """
+        :type n: int
+        :rtype: List[List[int]]
+        """
+        from math import sqrt
+        if n < 4:
+            return []
+        res = []
+        for i in range(2, int(sqrt(n))+1):
+            if n % i == 0 and i <= n / i:
+                res.append([i]+[n/i])
+                for j in self.getFactors(n/i):
+                    if j and i <= j[0]:
+                        res.append([i]+j)
+        return res
+```
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+
+想着用一下memorization会更快一点，果然beats 100%
+
+```python
+class Solution(object):
+    cache = {}
+    def getFactors(self, n):
+        """
+        :type n: int
+        :rtype: List[List[int]]
+        """
+        from math import sqrt
+        if n < 4:
+            return []
+        if n in self.cache:
+            return self.cache[n]
+        else:
+            res = []
+            for i in range(2, int(sqrt(n))+1):
+                if n % i == 0 and i <= n / i:
+                    res.append([i]+[n/i])
+                    for j in self.getFactors(n/i):
+                        if j and i <= j[0]:
+                            res.append([i]+j)
+            self.cache[n] = res
+        return res
+```