leetcode_Implement Queue using Stacks

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：

1.用栈来实现一个队列，也就是用后进先出的栈实现先进先出的队列

2.这个还是很难想的，但总之还是比用队列来实现栈容易想，大概就是用两个栈stack1和stack2来模拟队列

3.所有的元素都从stack1进栈，所有元素都从stack2出栈，当stack2为空的时候，将stack1中的所有元素出栈并全部push到stack2中去，然后从stack2出栈

4.由于栈是后进先出的，两次后进先出的操作就实现了队列的功能

代码：

class MyQueue {    Stack<Integer>stack1=new Stack<Integer>();Stack<Integer>stack2=new Stack<Integer>();// Push element x to the back of queue.    public void push(int x) {        stack1.push(x);    }    // Removes the element from in front of queue.    public void pop() {        if(!stack2.empty())        stack2.pop();        else         {while(!stack1.empty()){stack2.push(stack1.pop());}stack2.pop();}    }    // Get the front element.    public int peek() {    int num=0;    if(!stack2.empty())        num= stack2.peek();        else         {while(!stack1.empty()){stack2.push(stack1.pop());}num= stack2.peek();}        return num;    }    // Return whether the queue is empty.    public boolean empty() {        if(stack1.empty()&&stack2.empty())        return true;        return false;    }}