feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题16. 数值的整数次方

Author: yanglbme

lcof/面试题16. 数值的整数次方/README.md

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+# [面试题16. 数值的整数次方](https://leetcode-cn.com/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/)
+
+## 题目描述
+实现函数 double Power(double base, int exponent)，求 base 的 exponent 次方。不得使用库函数，同时不需要考虑大数问题。
+
+**示例 1:**
+
+```
+输入: 2.00000, 10
+输出: 1024.00000
+```
+
+**示例 2:**
+
+```
+输入: 2.10000, 3
+输出: 9.26100
+```
+
+**示例 3:**
+
+```
+输入: 2.00000, -2
+输出: 0.25000
+解释: 2-2 = 1/22 = 1/4 = 0.25
+```
+
+**说明:**
+
+- `-100.0 < x < 100.0`
+- n 是 32 位有符号整数，其数值范围是  `[−231, 231 − 1]` 。
+
+## 解法
+### Python3
+```python
+class Solution:
+    cache = {}
+    def myPow(self, x: float, n: int) -> float:
+        self.cache = {}
+        return self.pow(x, n)
+
+    def pow(self, x, n):
+        if self.cache.get('{}-{}'.format(x, n)):
+            return self.cache['{}-{}'.format(x, n)]
+        if n == 0: return 1
+        if n == 1: return x
+        if n == -1: return 1 / x
+
+        half = self.pow(x, n // 2)
+        self.cache['{}-{}'.format(x, n // 2)] = half
+        return half * half * self.pow(x, n % 2)
+```
+
+### Java
+```java
+class Solution {
+    public double myPow(double x, int n) {
+        if (n == 0) return 1;
+        if (n == 1) return x;
+        if (n == -1) return 1 / x;
+        double half = myPow(x, n / 2);
+        return half * half * myPow(x, n % 2);
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题16. 数值的整数次方/Solution.java

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+class Solution {
+    public double myPow(double x, int n) {
+        if (n == 0)
+            return 1;
+        if (n == 1)
+            return x;
+        if (n == -1)
+            return 1 / x;
+        double half = myPow(x, n / 2);
+        return half * half * myPow(x, n % 2);
+    }
+}

lcof/面试题16. 数值的整数次方/Solution.py

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+class Solution:
+    cache = {}
+    def myPow(self, x: float, n: int) -> float:
+        self.cache = {}
+        return self.pow(x, n)
+
+    def pow(self, x, n):
+        if self.cache.get('{}-{}'.format(x, n)):
+            return self.cache['{}-{}'.format(x, n)]
+        if n == 0: return 1
+        if n == 1: return x
+        if n == -1: return 1 / x
+
+        half = self.pow(x, n // 2)
+        self.cache['{}-{}'.format(x, n // 2)] = half
+        return half * half * self.pow(x, n % 2)