面试题 03.02. 栈的最小值

请设计一个栈，除了常规栈支持的pop与push函数以外，还支持min函数，该函数返回栈元素中的最小值。执行push、pop和min操作的时间复杂度必须为O(1)。

示例：

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); –> 返回 -3.
minStack.pop();
minStack.top(); –> 返回 0.
minStack.getMin(); –> 返回 -2.

Python 解答：

class MinStack:

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

    def push(self, x: int) -> None:
        self.Stack.append(x)
        if not self.MinS:
            self.MinS.append(x)
        elif x <= self.MinS[-1]:
            self.MinS.append(x)

    def pop(self) -> None:
        if self.Stack:
            temp = self.Stack.pop()
            if self.MinS[-1] == temp:
                self.MinS.pop()

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

    def getMin(self) -> int:
        if self.Stack:
            return self.MinS[-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()