Create medium-Pow(x, n).md

leetcode/medium-Pow(x, n).md

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+Pow(x, n) (Leetcode #50)
+===============================
+### Medium
+
+Implement `pow(x, n)`, which calculates `x` raised to the power `n` (i.e., `x^n`).
+
+
+
+### Example 1:
+```
+Input: x = 2.00000, n = 10
+Output: 1024.00000
+```
+
+### Example 2:
+```
+Input: x = 2.10000, n = 3
+Output: 9.26100
+```
+
+### Example 3:
+```
+Input: x = 2.00000, n = -2
+Output: 0.25000
+Explanation: 2-2 = 1/22 = 1/4 = 0.25
+ ```
+
+
+### Constraints:
+```
+-100.0 < x < 100.0
+-2^31 <= n <= 2^31-1
+n is an integer.
+Either x is not zero or n > 0.
+-10^4 <= xn <= 10^4
+```
+
+### Solution
+```python
+# T: O(log n)
+class Solution:
+    def myPow(self, x: float, n: int) -> float:
+        if n == 0 or x == 1:
+            return 1
+        if x == -1:
+            return -1 if n%2 else 1
+
+        x = x if n > 0 else 1/x
+        ans = x
+        n = abs(n)
+        p = 1
+        while p*p < n:
+            p *= 2
+            ans = ans*ans
+            if ans < 1e-6:
+                return 0
+
+        for _ in range(p, n):
+            ans = ans*x
+        return ans
+
+## recursive solution
+# class Solution:
+#     def myPow(self, x: float, n: int) -> float:
+#         if abs(x) < 1e-9:
+#             return 0            
+#         if n == 0 or x == 1:
+#             return 1
+
+#         if n < 0:
+#             return self.myPow(1/x, -n)
+
+#         level = self.myPow(x, n//2)
+#         if n % 2:
+#             return level*level*x
+#         else:
+#             return level*level
+
+```
+
+C++
+```c++
+class Solution {
+public:
+    double myPow(double x, int n) {        
+        if (abs(x) < 1e-9)
+            return 0;
+        if (n==0 || x==1)
+            return 1.;
+
+        if (n < 0) {
+            if (n == INT_MIN)
+                return myPow(1/x, -(n+1))/x;
+            else
+                return myPow(1/x, -n);
+        }
+        double level = myPow(x, floor(n/2));
+        if (n % 2)
+            return level*level*x;
+        else
+            return level*level;
+    }
+};
+```