>>> stack = [3, 4, 5]
>>> stack.append(6)
>>> stack.append(7)
>>> stack
[3, 4, 5, 6, 7]
>>> stack.pop()
7
>>> stack
[3, 4, 5, 6]
>>> stack.pop()
6
>>> stack.pop()
5
>>> stack
[3, 4]

佇列

>>> from collections import
 
 deque
>>> queue = deque(["Eric", "John", "Michael"])
>>> queue.append("Terry")           # Terry arrives
>>> queue.append("Graham")          # Graham arrives
>>> queue.popleft()                 # The first to arrive now leaves
'Eric'
>>> queue.popleft()                 # The second to arrive now leaves
 

'John'
>>> queue                           # Remaining queue in order of arrival
deque(['Michael', 'Terry', 'Graham'])

使用兩個棧實現一個佇列 插入：插入到stack1 刪除：如果stack2為空，將stack1pop之後push到stack2，如果stack2不為空，則直接popstack2

class Solution:
    def __init__(self):
        self.stack1 = []
        self.stack2 =
 
 []
    def push(self, node):
        self.stack1.append(node)
    def pop(self):
        if self.stack2 == []:
            while self.stack1 != []:
                self.stack2.append(self.stack1.pop())
        return self.stack2.pop()
sol = Solution()
sol.push(1)
sol.push(2)
print(sol.pop())
sol.push(3)
print(sol.pop())
print(sol.pop())

輸出：

1
2
3