feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题30. 包含min函数的栈

Author: yanglbme

lcof/面试题30. 包含min函数的栈/README.md

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+# [面试题30. 包含min函数的栈](https://leetcode-cn.com/problems/bao-han-minhan-shu-de-zhan-lcof/)
+
+## 题目描述
+定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
+
+**示例:**
+
+```
+MinStack minStack = new MinStack();
+minStack.push(-2);
+minStack.push(0);
+minStack.push(-3);
+minStack.min();   --> 返回 -3.
+minStack.pop();
+minStack.top();      --> 返回 0.
+minStack.min();   --> 返回 -2.
+```
+
+**提示：**
+
+- 各函数的调用总次数不超过 20000 次
+
+## 解法
+### Python3
+```python
+class MinStack:
+
+    def __init__(self):
+        """
+        initialize your data structure here.
+        """
+        self._s1 = []
+        self._s2 = []
+
+    def push(self, x: int) -> None:
+        self._s1.append(x)
+        self._s2.append(x if len(self._s2) == 0 or self._s2[-1] > x else self._s2[-1])
+
+    def pop(self) -> None:
+        self._s1.pop()
+        self._s2.pop()
+
+    def top(self) -> int:
+        return self._s1[-1]
+
+    def min(self) -> int:
+        return self._s2[-1]
+
+
+# Your MinStack object will be instantiated and called as such:
+# obj = MinStack()
+# obj.push(x)
+# obj.pop()
+# param_3 = obj.top()
+# param_4 = obj.min()
+```
+
+### Java
+```java
+class MinStack {
+    private Stack<Integer> s1;
+    private Stack<Integer> s2;
+    
+    /** initialize your data structure here. */
+    public MinStack() {
+        s1 = new Stack<>();
+        s2 = new Stack<>();
+    }
+    
+    public void push(int x) {
+        s1.push(x);
+        s2.push((s2.empty() || s2.peek() > x) ? x : s2.peek());
+    }
+    
+    public void pop() {
+        s1.pop();
+        s2.pop();
+    }
+    
+    public int top() {
+        return s1.peek();
+    }
+    
+    public int min() {
+        return s2.peek();
+    }
+}
+
+/**
+ * Your MinStack object will be instantiated and called as such:
+ * MinStack obj = new MinStack();
+ * obj.push(x);
+ * obj.pop();
+ * int param_3 = obj.top();
+ * int param_4 = obj.min();
+ */
+```
+
+### ...
+```
+
+```

lcof/面试题30. 包含min函数的栈/Solution.java

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+class MinStack {
+    private Stack<Integer> s1;
+    private Stack<Integer> s2;
+    
+    /** initialize your data structure here. */
+    public MinStack() {
+        s1 = new Stack<>();
+        s2 = new Stack<>();
+    }
+    
+    public void push(int x) {
+        s1.push(x);
+        s2.push((s2.empty() || s2.peek() > x) ? x : s2.peek());
+    }
+    
+    public void pop() {
+        s1.pop();
+        s2.pop();
+    }
+    
+    public int top() {
+        return s1.peek();
+    }
+    
+    public int min() {
+        return s2.peek();
+    }
+}
+
+/**
+ * Your MinStack object will be instantiated and called as such:
+ * MinStack obj = new MinStack();
+ * obj.push(x);
+ * obj.pop();
+ * int param_3 = obj.top();
+ * int param_4 = obj.min();
+ */

lcof/面试题30. 包含min函数的栈/Solution.py

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+class MinStack:
+
+    def __init__(self):
+        """
+        initialize your data structure here.
+        """
+        self._s1 = []
+        self._s2 = []
+
+    def push(self, x: int) -> None:
+        self._s1.append(x)
+        self._s2.append(x if len(self._s2) == 0 or self._s2[-1] > x else self._s2[-1])
+
+    def pop(self) -> None:
+        self._s1.pop()
+        self._s2.pop()
+
+    def top(self) -> int:
+        return self._s1[-1]
+
+    def min(self) -> int:
+        return self._s2[-1]
+
+
+# Your MinStack object will be instantiated and called as such:
+# obj = MinStack()
+# obj.push(x)
+# obj.pop()
+# param_3 = obj.top()
+# param_4 = obj.min()