LeetCode:Implement Queue using Stacks

问题描述：

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.

Notes:

• You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is emptyoperations 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).

思路：

利用两个栈实现一个队列，模拟队列“先进先出”思想，stack1为入栈，stack2为出栈，入队直接进入stack1即可；出队时，如果stack2有元素，直接是stack2出栈，如果没有元素，则先将stack1的元素放入stack2中，这时最上面的元素出栈即可；获取最顶元素和出队类似；判断队列是否为空时，必须判断两个栈是否都为空才可以。

JAVA代码：

class MyQueue {
    // Push element x to the back of queue.
    Stack<Integer> stack1 = new Stack<>();
    Stack<Integer> stack2 = new Stack<>();
    
    public void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        if(!stack2.isEmpty())   stack2.pop();  
        else{ 
            while(!stack1.isEmpty())
                stack2.push(stack1.pop());
            stack2.pop();
        }
    }

    // Get the front element.
    public int peek() {
        if(!stack2.isEmpty())   return stack2.peek();
        else{
            while(!stack1.isEmpty())
                stack2.push(stack1.pop());
            return stack2.peek();
        }
    }

    // Return whether the queue is empty.
    public boolean empty() 
    {
         return  stack1.empty() && stack2.empty();  
    }
}