## Description

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.


## Solutions

### 1. 两个 Stack

class MinStack:

def __init__(self):
"""
"""
self.stack = []
self.min = []

def push(self, x: int) -> None:
if len(self.min) == 0 or x <= self.min[-1]:
self.min.append(x)
self.stack.append(x)

def pop(self) -> None:
if self.stack[-1] == self.min[-1]:
self.min.pop()
self.stack.pop()

def top(self) -> int:
return self.stack[-1]

def getMin(self) -> int:
return self.min[-1]
# Runtime: 68 ms, faster than 91.20% of Python3 online submissions for Min Stack.
# Memory Usage: 17.5 MB, less than 5.36% of Python3 online submissions for Min Stack.

# 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()