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/

posted on 2019-10-22 10:01  星辰之衍  阅读(49)  评论(0编辑  收藏

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