國際慣例，先上題目：

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 useonlystandard operations of a stack -- which means onlypush to top,peek/pop from top,size, andis 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).

這道題個人覺得屬于比較有意思的一道題><還有一道類似的用queue來完成stack的，不過我還沒有做到，暫且不說。

這道題比較好想到的方法是雙stack法，因為剛好stack和queue的順序是相反的。

試想建立兩個stack，當把所有數據從stack a中轉移到b的時候，其數據排列就會發生變化，這樣一來就很好理解了。

不過也可以只建立一個stack，這里就需要swap來輔助，這個會比較節省時間一點。

不過我暫時只寫了雙stack的用法，下次有機會寫下單stack的哈哈。

個人覺得這道題難點就在第一個method上，后面都很簡單，直接套用stack的method即可。

class MyQueue {        Stack<Integer> a=new Stack<Integer>();    Stack<Integer> b=new Stack<Integer>();        // Push element x to the back of queue.    public void push(int x) {        //specail case        if(a.isEmpty()){            a.push(x);        }else{            while(!a.isEmpty()){                b.push(a.pop());            }            a.push(x);            while(!b.isEmpty()){                a.push(b.pop());            }        }            }    // Removes the element from in front of queue.    public void pop() {        a.pop();    }    // Get the front element.    public int peek() {        return a.peek();    }    // Return whether the queue is empty.    public boolean empty() {        return a.isEmpty();    }}