---
description: '# easy'
---

# 155. Min Stack

{% hint style="info" %}
It's easy to consider to use two stacks, one stack is the normal one, the other stack stores the minimum elements.

key idea: min\_stack - 存储push to stack目前为止的最小值，所以可能有很多值，易错点：在push时，即便这个数大于等于min\_stack.top()的数，也要压入min\_stack
{% endhint %}

```python
class MinStack:

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

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

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

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

    def getMin(self) -> int:
        return self.min_stack[-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()
```

Time = $$O(1)$$ , space = $$O(n)$$&#x20;
