[LeetCode] 232. Implement Queue using Stacks_Easy tag: Design

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, and is 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).

这个题目可以用两个stack, 基本思路是如@elmirap提出来的Solution, 两种方式, 第一种是O(n) push, 其他都是O(1)

Code:

class MyQueue(object):
    ##Solution
    # two stacks, O(n) push operation

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

    def push(self, x):
        """
        Push element x to the back of queue.
        :type x: int
        :rtype: void
        """
        while self.stack:
            self.backup.append(self.stack.pop())
        self.stack.append(x)
        while self.backup:
            self.stack.append(self.backup.pop())
        

    def pop(self):
        """
        Removes the element from in front of queue and returns that element.
        :rtype: int
        """
        return self.stack.pop()
        

    def peek(self):
        """
        Get the front element.
        :rtype: int
        """
        return self.stack[-1]
        

    def empty(self):
        """
        Returns whether the queue is empty.
        :rtype: bool
        """
        return not self.stack
        


# 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()

 

 

第二种方式, 平均pop为O(1), 其他都为O(1), 思路实际上类似, 但是在Stack2不急着将其搬回stack1, 因为在stack1搬到stack2之后已经将最后的放在最下面了.所以节约了时间.

 

Code:

class MyQueue:

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

    def push(self, x):
        """
        Push element x to the back of queue.
        :type x: int
        :rtype: void
        """
        self.stack.append(x)
        

    def pop(self):
        """
        Removes the element from in front of queue and returns that element.
        :rtype: int
        """
        if not self.queue:
            while self.stack:
                self.queue.append(self.stack.pop())
        return self.queue.pop()
        
            
        

    def peek(self):
        """
        Get the front element.
        :rtype: int
        """
        return self.queue[-1] if self.queue else self.stack[0]
        

    def empty(self):
        """
        Returns whether the queue is empty.
        :rtype: bool
        """
        return not self.stack and not self.queue
        


# 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()