Design a stack that supports getMin() in O(1) time and O(1) extra space

Question: Design a Data Structure SpecialStack that supports all the stack operations like push(), pop(), isEmpty(), isFull() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must be O(1). To implement SpecialStack, you should only use standard Stack data structure and no other data structure like arrays, list, .. etc.

Example:

Consider the following SpecialStack
16  --> TOP
15
29
19
18

When getMin() is called it should return 15,
which is the minimum element in the current stack. 

If we do pop two times on stack, the stack becomes
29  --> TOP
19
18

When getMin() is called, it should return 18
which is the minimum in the current stack.

• 解答

# Class to make a Node
class Node:
	# Constructor which assign argument to nade's value
	def __init__(self, value):
		self.value = value
		self.next = None

	# This method returns the string representation of the object.
	def __str__(self):
		return "Node({})".format(self.value) 

	# __repr__ is same as __str__
	__repr__ = __str__ 

class Stack:
	# Stack Constructor initialise top of stack and counter.
	def __init__(self):
		self.top = None
		self.count = 0
		self.minimum = None

	# This method returns the string representation of the object (stack).
	def __str__(self):
		temp = self.top
		out = []
		while temp:
			out.append(str(temp.value))
			temp = temp.next
		out = '\n'.join(out)
		return ('Top {} \n\nStack :\n{}'.format(self.top,out)) 

	# __repr__ is same as __str__
	__repr__=__str__ 

	# This method is used to get minimum element of stack
	def getMin(self):
		if self.top is None:
			return "Stack is empty"
		else:
			print("Minimum Element in the stack is: {}" .format(self.minimum)) 

	# Method to check if Stack is Empty or not
	def isEmpty(self):
		# If top equals to None then stack is empty
		if self.top == None:
			return True
		else:
		# If top not equal to None then stack is empty
			return False

	# This method returns length of stack
	def __len__(self):
		self.count = 0
		tempNode = self.top
		while tempNode:
			tempNode = tempNode.next
			self.count+=1
		return self.count 

	# This method returns top of stack
	def peek(self):
		if self.top is None:
			print ("Stack is empty")
		else:
			if self.top.value < self.minimum:
				print("Top Most Element is: {}" .format(self.minimum))
			else:
				print("Top Most Element is: {}" .format(self.top.value)) 

	# This method is used to add node to stack
	def push(self,value):
		if self.top is None:
			self.top = Node(value)
			self.minimum = value 

		elif value < self.minimum:
			temp = (2 * value) - self.minimum
			new_node = Node(temp)
			new_node.next = self.top
			self.top = new_node
			self.minimum = value
		else:
			new_node = Node(value)
			new_node.next = self.top
			self.top = new_node
		print("Number Inserted: {}" .format(value)) 

	# This method is used to pop top of stack
	def pop(self):
		if self.top is None:
			print( "Stack is empty")
		else:
			removedNode = self.top.value
			self.top = self.top.next
			if removedNode < self.minimum:
				print ("Top Most Element Removed :{} " .format(self.minimum))
				self.minimum = ( ( 2 * self.minimum ) - removedNode )
			else:
				print ("Top Most Element Removed : {}" .format(removedNode)) 

# Driver program to test above class
stack = Stack() 

stack.push(3)
stack.push(5)
stack.getMin()
stack.push(2)
stack.push(1)
stack.getMin()
stack.pop()
stack.getMin()
stack.pop()
stack.peek() 

# This code is contributed by Blinkii

• 分析

https://www.geeksforgeeks.org/design-a-stack-that-supports-getmin-in-o1-time-and-o1-extra-space/