Leetcode +155 Min Stack

Leetcode +155 Min Stack
タイトルの説明
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.

ソース:力ボタン(LeetCode)リンク:https://leetcode-cn.com/problems/min-stack著作権はインターネットの所有に帰属する.商業転載は公式の授権に連絡してください.非商業転載は出典を明記してください.
構想解析

class MinStack:
    def __init__(self):
        #    
        self.data = []
        #    
        self.helper = []

    def push(self, x):
        self.data.append(x)
        if len(self.helper) == 0 or x <= self.helper[-1]:
            self.helper.append(x)
        else:
            self.helper.append(self.helper[-1])

    def pop(self):
        if self.data:
            self.helper.pop()
            return self.data.pop()

    def top(self):
        if self.data:
            return self.data[-1]

    def getMin(self):
        if self.helper:
            return self.helper[-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()