LeetCode155. Min Stack

常数时间复杂度的栈与最小值操作

最新推荐文章于 2022-08-31 23:43:21 发布

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本文介绍了一种特殊的栈实现，该栈除了支持常规的push、pop和top操作外，还能在常数时间内检索栈中最小元素。通过实例演示了如何使用这种栈，包括初始化、元素压入、弹出、获取顶部元素及最小值。

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.

<思路>写一个栈的功能。多了一个获取最小值操作。float('inf')表示正无穷。需要注意的是在pop()后要更新最小值。

class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.min = float('inf')

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        self.stack.append(x)
        if x < self.min:
            self.min = x
        

    def pop(self):
        """
        :rtype: void
        """
        self.stack.pop()
        if self.stack:
            self.min = min(self.stack)
        else:
            self.min = float('inf')
    def top(self):
        """
        :rtype: int
        """
        return self.stack[-1]

    def getMin(self):
        """
        :rtype: int
        """
        return self.min


# 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()