Implement Queue using Stacks

mplement 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.

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).

这是＜剑指offer>上的一道题．与Implement Stack using Queues互为对偶．这里主要的难点不在添加元素，而在于如何利用从尾部pop的栈实现从头pop().使用栈没法像queue那样进行左旋，所以不可避免要使用两个stack,stack1为input栈，stack2为ouput栈．input栈负责增加元素，一旦要进行peek或者pop操作，output栈就要启用，如果为空的话，需要从input栈中逆序增加元素．因为queue先进先出，所以后来再加入stack1的元素，在stack2不为空的情况下，并不影响peek和pop操作．代码如下：

class Queue(object):
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []
        
    def adjust(self):
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
                
    def push(self, x):
        """
        :type x: int
        :rtype: nothing
        """
        self.stack1.append(x)
        

    def pop(self):
        """
        :rtype: nothing
        """
        if not self.empty():
            self.stack2.pop()
        

    def peek(self):
        """
        :rtype: int
        """
        if not self.empty():
            return self.stack2[-1]
        

    def empty(self):
        """
        :rtype: bool
        """
        self.adjust()
        return len(self.stack2) == 0 

给出一个leetcode上的参考代码，和我的没有本质差别：

class Queue {
    stack<int> input, output;
public:

    void push(int x) {
        input.push(x);
    }

    void pop(void) {
        peek();
        output.pop();
    }

    int peek(void) {
        if (output.empty())
            while (input.size())
                output.push(input.top()), input.pop();
        return output.top();
    }

    bool empty(void) {
        return input.empty() && output.empty();
    }
};

 每个元素都会在stack1,stack2存储一次，所以所有操作的均摊复杂度都为Ｏ(1)．

转载于:https://www.cnblogs.com/sherylwang/p/5595906.html