面试题30. 包含min函数的栈

定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。

示例:

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

来源：力扣（LeetCode）

思路：用辅助栈存储当前data的最小值，辅助栈头即为min值

class MinStack:

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


    def push(self, x: int) -> None:
        self.stack.append(x)
        
        if not self.min_stack:
            self.min_stack.append(x)
        else:
            if self.min_stack[-1] < x:
                self.min_stack.append(self.min_stack[-1])
            else:
                self.min_stack.append(x)


    def pop(self) -> None:
       self.stack.pop(-1)
       self.min_stack.pop(-1)



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


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


# 用辅助栈存储当前data的最小值，辅助栈头即为min值

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