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