leetcode 232. Implement Queue using Stacks
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Implement the following operations of a queue using stacks.
- push(x) -- Push element x to the back of queue.
- pop() -- Removes the element from in front of queue.
- peek() -- Get the front element.
- empty() -- Return whether the queue is empty.
Example:
MyQueue queue = new MyQueue(); queue.push(1); queue.push(2); queue.peek(); // returns 1 queue.pop(); // returns 1 queue.empty(); // returns false
Notes:
- You must use only standard operations of a stack -- which means only
push to top
,peek/pop from top
,size
, andis empty
operations are valid. - Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
class MyQueue: def __init__(self): """ Initialize your data structure here. """ self.stack1 = [] self.stack2 = [] def push(self, x): """ Push element x to the back of queue. :type x: int :rtype: void """ self.stack1.append(x) def dump_data(self): if not self.stack2: while self.stack1: self.stack2.append(self.stack1.pop()) def pop(self): """ Removes the element from in front of queue and returns that element. :rtype: int """ self.dump_data() return self.stack2.pop() def peek(self): """ Get the front element. :rtype: int """ self.dump_data() return self.stack2[-1] def empty(self): """ Returns whether the queue is empty. :rtype: bool """ return not self.stack1 and not self.stack2 # 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()
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