最小栈 一种解法(Python)

使用辅助栈 时间复杂度 O(1)

非同步栈

class MinStack:

    def __init__(self):
        self.stack = []
        self.min_stack = []


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

    def pop(self) -> None:
        if self.stack.pop() == self.min_stack[-1]:
            self.min_stack.pop()


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

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

同步栈

class MinStack:

    def __init__(self):
        self.stack = []
        # 添加正无穷 这样第一次添加 不用判断边界
        self.min_stack = [math.inf]


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

    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]