[Lintcode] 40. Implement Queue by Two Stacks/[Leetcode]225. Implement Stack using Queues

40. Implement Queue by Two Stacks / 225. Implement Stack using Queues

• 本题难度: Medium/Easy

• Topic: Data Structure - stack/queue

  Description

  As the title described, you should only use two stacks to implement a queue's actions.

The queue should support push(element), pop() and top() where pop is pop the first(a.k.a front) element in the queue.

Both pop and top methods should return the value of first element.

Example
Example 1:

Input:
push(1)
pop()
push(2)
push(3)
top()
pop()
Output:
1
2
2

Example 2:

Input:
push(1)
push(2)
push(2)
push(3)
push(4)
push(5)
push(6)
push(7)
push(1)
Output:
[]

Challenge
implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE.

我的代码

class MyQueue:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []
        

    def push(self, x: 'int') -> 'None':
        """
        Push element x to the back of queue.
        """
        while(len(self.stack1)>0):
            self.stack2.append(self.stack1.pop())
        self.stack1.append(x)
        while(len(self.stack2)>0):
            self.stack1.append(self.stack2.pop())
        return 
        

    def pop(self) -> 'int':
        """
        Removes the element from in front of queue and returns that element.
        """
        return self.stack1.pop()
        

    def peek(self) -> 'int':
        """
        Get the front element.
        """
        return self.stack1[-1]
        

    def empty(self) -> 'bool':
        """
        Returns whether the queue is empty.
        """
        return self.stack1 == []
        


# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()

思路
example:

push 2

list1 [2]
list2 []

push 3
list1[2]       ->   list1 [3]   ->      list1 [3,2]
list2[]        ->   list2 [2]   ->      list2 []

push 4
list1 [3,2]    -> list1[4]      ->      list1[4,3,2]
list2 []       -> lst2[2,3]     ->      list2[]

• 时间复杂度 push O(n) pop, peek, empty O(1)

转载于:https://www.cnblogs.com/siriusli/p/10392464.html