5/30 每日一題(定義堆疊,使用兩個stack實現)

 Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

• You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

 

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

 

Constraints:

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid.

 

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

class MyQueue:

#要求使用2個stack模擬出先進先出(FIFO)

#第一個stack1紀錄元素的推入情況

#第二個stack2紀錄元素的彈出情況

#要push時使用stack1 ,當stack2不為空時彈出stack2元素

    def __init__(self):

        self.stack1=list()

        self.stack2=list()

    def push(self, x: int) -> None:

        #stack1是用來記錄元素推入順序的堆疊

        self.stack1.append(x)

       

       

    def pop(self) -> int:

        #stack1=[1,2] >>>stack2=[2,1] #彈出stack2即滿足FIFO

        #將stack1的值彈入到stack2是為了確保彈出順序FIFO

        if not self.stack2:            

            while self.stack1:

                self.stack2.append(self.stack1.pop())

        return self.stack2.pop()

       

    def peek(self) -> int:  

        #如果stack2存在,那最初進入的元素就是stack2的最後一項    

        if self.stack2:

            return self.stack2[-1]

        else:

            while self.stack1:

                self.stack2.append(self.stack1.pop())

            return self.stack2[-1]                

       

    def empty(self) -> bool:

        #當同時滿足 stack1 且  stack2 都為空時,就表示為他是empty

        return len(self.stack1)==0 and len(self.stack2)==0

       

# 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() ----------------------------------------參考解答

class MyQueue(object):
    def __init__(self):
        self.in_stk = []
        self.out_stk = []
	# Push element x to the back of queue...
    def push(self, x):
        self.in_stk.append(x)
	# Remove the element from the front of the queue and returns it...
    def pop(self):
        self.peek()
        return self.out_stk.pop()
	# Get the front element...
    def peek(self):
        if not self.out_stk:
            while self.in_stk:
                self.out_stk.append(self.in_stk.pop())
        return self.out_stk[-1]
	# Return whether the queue is empty...
    def empty(self):
        return not self.in_stk and not self.out_stk