Min Stack

题目描述

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.

解题方法

• 用另一个stack存可能的最小值minStack
• 如果新Push的value大于当前的最小值，就不用存，只有当新的value <= current min时, 才push到这个minStack上
• 如果pop的value == minStack的top, 才从它pop

Solution

class MinStack(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.minStack = []

    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.stack.append(x)
        if not self.minStack or x <= self.minStack[-1]:
            self.minStack.append(x)

    def pop(self):
        """
        :rtype: nothing
        """
        ele = self.stack.pop()
        if ele == self.minStack[-1]:
            self.minStack.pop()

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

    def getMin(self):
        """
        :rtype: int
        """
        return self.minStack[-1]

Reference