feat: add python and java solution to lcof problem

Author: yanglbme

lcof/README.md

@@ -26,7 +26,15 @@
 │   ├── README.md
 │   ├── Solution.java
 │   └── Solution.py
-└── 面试题09. 用两个栈实现队列
+├── 面试题09. 用两个栈实现队列
+│   ├── README.md
+│   ├── Solution.java
+│   └── Solution.py
+├── 面试题10- I. 斐波那契数列
+│   ├── README.md
+│   ├── Solution.java
+│   └── Solution.py
+└── 面试题10- II. 青蛙跳台阶问题
     ├── README.md
     ├── Solution.java
     └── Solution.py

lcof/面试题10- II. 青蛙跳台阶问题/README.md

@@ -0,0 +1,57 @@
+# [面试题10- II. 青蛙跳台阶问题](https://leetcode-cn.com/problems/qing-wa-tiao-tai-jie-wen-ti-lcof/)
+
+## 题目描述
+一只青蛙一次可以跳上 1 级台阶，也可以跳上 2 级台阶。求该青蛙跳上一个 `n` 级的台阶总共有多少种跳法。
+
+答案需要取模 1e9+7（1000000007），如计算初始结果为：1000000008，请返回 1。
+
+**示例 1：**
+
+```
+输入：n = 2
+输出：2
+```
+
+**示例 2：**
+
+```
+输入：n = 7
+输出：21
+```
+
+**提示：**
+
+- `0 <= n <= 100`
+
+
+## 解法
+### Python3
+```python
+class Solution:
+    def numWays(self, n: int) -> int:
+        a, b = 0, 1
+        for _ in range(n):
+            s = a + b
+            a, b = b, s
+        return b % 1000000007
+```
+
+### Java
+```java
+class Solution {
+    public int numWays(int n) {
+        int a = 0, b = 1;
+        for (int i = 0; i < n; ++i) {
+            int s = (a + b) % 1000000007;
+            a = b;
+            b = s;
+        }
+        return b;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题10- II. 青蛙跳台阶问题/Solution.java

@@ -0,0 +1,11 @@
+class Solution {
+    public int numWays(int n) {
+        int a = 0, b = 1;
+        for (int i = 0; i < n; ++i) {
+            int s = (a + b) % 1000000007;
+            a = b;
+            b = s;
+        }
+        return b;
+    }
+}

lcof/面试题10- II. 青蛙跳台阶问题/Solution.py

@@ -0,0 +1,7 @@
+class Solution:
+    def numWays(self, n: int) -> int:
+        a, b = 0, 1
+        for _ in range(n):
+            s = a + b
+            a, b = b, s
+        return b % 1000000007