leetcode | 155.最小栈

155.最小栈

题目描述

设计一个支持 push ，pop ，top 操作，并能在常数时间内检索到最小元素的栈。
push(x) —— 将元素 x 推入栈中。
pop() —— 删除栈顶的元素。
top() —— 获取栈顶元素。
getMin() —— 检索栈中的最小元素。

题目链接

思路

栈的性质：先进后出

代码

class MinStack(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = list()

    def push(self, val):
        """
        :type val: int
        :rtype: None
        """
        self.stack.append(val)

    def pop(self):
        """
        :rtype: None
        """
        if len(self.stack) != 0:
            self.stack.pop()

    def top(self):
        """
        :rtype: int
        """
        if len(self.stack) != 0:
            return self.stack[-1]

    def getMin(self):
        """
        :rtype: int
        """
        if len(self.stack) != 0:
            return min(self.stack)

# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()

import math

class MinStack1:
    """辅助栈"""
    def __init__(self):
        self.stack = []
        self.min_stack = [math.inf]

    def push(self, x: int) -> None:
        self.stack.append(x)
        self.min_stack.append(min(x, self.min_stack[-1]))

    def pop(self) -> None:
        self.stack.pop()
        self.min_stack.pop()

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

    def getMin(self) -> int:
        return self.min_stack[-1]



if __name__ == "__main__":
    # min_stack = MinStack()
    min_stack = MinStack1()
    min_stack.push(-2)
    min_stack.push(0)
    min_stack.push(-3)
    print(min_stack.getMin())
    min_stack.pop()
    print(min_stack.top())
    print(min_stack.getMin())
    print(math.inf)