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

标签：bec   com   structure   should   sign   stand   ons   min()   return   

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/

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

标签：bec   com   structure   should   sign   stand   ons   min()   return   

原文地址：https://www.cnblogs.com/HongjianChen/p/11607454.html