# Min Stack

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

* push(x) -- Push element x onto stack.
* pop() -- Removes the element on top of the stack.
* top() -- Get the top element.
* getMin() -- Retrieve the minimum element in the stack.

**Example:**

```
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.
```

## 题目大意：

设计一个栈，支持在**常数时间**内push，pop，top，和取最小值。

* push(x) -- 元素x压入栈
* pop() -- 弹出栈顶元素
* top() -- 获取栈顶元素
* getMin() -- 获取栈中的最小值

## 解题思路：

“双栈法”，栈stack存储当前的所有元素，minStack存储栈中的最小元素。

在操作元素栈stack的同时，维护最小值栈minStack。

```
class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.node = []
        self.minnode =[]


    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        self.node.append(x)
        if self.minnode:
            x = min(self.minnode[-1], x)
        self.minnode.append(x)


    def pop(self):
        """
        :rtype: void
        """
        self.minnode.pop()
        self.node.pop()



    def top(self):
        """
        :rtype: int
        """
        return self.node[-1]


    def getMin(self):
        """
        :rtype: int
        """
        return self.minnode[-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.getMin()
```


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