[leetcode]716. Max Stack @ python

版权声明：版权归个人所有，未经博主允许，禁止转载 https://blog.csdn.net/danspace1/article/details/88734584

原题

Design a max stack that supports push, pop, top, peekMax and popMax.

push(x) – Push element x onto stack.
pop() – Remove the element on top of the stack and return it.
top() – Get the element on the top.
peekMax() – Retrieve the maximum element in the stack.
popMax() – Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one.
Example 1:
MaxStack stack = new MaxStack();
stack.push(5);
stack.push(1);
stack.push(5);
stack.top(); -> 5
stack.popMax(); -> 5
stack.top(); -> 1
stack.peekMax(); -> 5
stack.pop(); -> 1
stack.top(); -> 5
Note:
-1e7 <= x <= 1e7
Number of operations won’t exceed 10000.
The last four operations won’t be called when stack is empty.

解法

使用列表, 每次push时向列表左边添加元素.

代码

class MaxStack(object):

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

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        self.data.insert(0, x)

    def pop(self):
        """
        :rtype: int
        """
        return self.data.pop(0)

    def top(self):
        """
        :rtype: int
        """
        return self.data[0]

    def peekMax(self):
        """
        :rtype: int
        """
        return max(self.data)

    def popMax(self):
        """
        :rtype: int
        """
        ans = max(self.data)
        self.data.remove(ans)
        return ans


# Your MaxStack object will be instantiated and called as such:
# obj = MaxStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.peekMax()
# param_5 = obj.popMax()