2019/8/3: Minimum stack

Problem description

Design a stack that supports push, pop, top operations and can retrieve the smallest elements in a constant time.

Source: LeetCode
Links: https://leetcode-cn.com/problems/min-stack
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Complete code

class MinStack:

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

    def push(self, x: int) -> None:
        self.item.append(x)

    def pop(self) -> None:
        if len(self.item) > 0:
            return self.item.pop()
        else:
            return 0

    def top(self) -> int:
        if len(self.item) != "":
            return self.item[len(self.item)-1]
        else:
            return 0

    def getMin(self) -> int:
        if len(self.item) > 0:
            return min(self.item)
        else:
            return 0


# 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()

Well, it feels like a little bit of a problem. After referring to other people's code, we find that we need to use two lists to achieve it. Because the minimum value of the stack is required to be retrieved in a constant time, and min() takes a long time, we introduce minStack list to store the minimum value:

class MinStack(object):
 
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []       
        self.minStack = []
 
    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        self.stack.append(x)
        if not self.minStack or self.minStack[-1]>=x:
            self.minStack.append(x)
 
    def pop(self):   
        """
        :rtype: void
        """
        if self.minStack[-1]==self.stack[-1]:
            del self.minStack[-1]
        self.stack.pop()
 
    def top(self):
        """
        :rtype: int
        """
        return self.stack[-1]
        
    def getMin(self):
        """
        :rtype: int
        """
        return self.minStack[-1]