Implement Stack using Queues

题目描述：

Implement the following operations of a stack using queues.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
empty() -- Return whether the stack is empty.
Notes:
You must use only standard operations of a queue -- which means only 

push
to back

, 

peek/pop from front

, 

size

,
and 

is empty

 operations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

解题思路：
使用两个队列，时刻保持一个队列为空。

push：push操作时直接把数值x push到非空的队列；

pop：把非空队列中的元素依次出队列直到队列中剩下一个元素，并把出队列的元素依次push到空队列中，这样院非空队列中剩下的元素就是栈顶元素，直接pop即可

top：把非空队列中的元素依次出队列直到队列中剩下一个元素，并把出队列的元素依次push到空队列中，这样院非空队列中剩下的元素就是栈顶元素，直接front即可，然后把该元素在push到原空队列中

empty：判断两个队列是否都为空即可

AC代码如下：

class Stack {
public:
// Push element x onto stack.
void push(int x) {
if (!que1.empty()){
que1.push(x);
}
else{
que2.push(x);
}
}

// Removes the element on top of the stack.
void pop() {
if (!que1.empty()){
while (que1.size() > 1){
que2.push(que1.front());
que1.pop();
}
que1.pop();
}
else{
while (que2.size() > 1){
que1.push(que2.front());
que2.pop();
}
que2.pop();
}
}

// Get the top element.
int top() {
int ans;
if (!que1.empty()){
while (que1.size() > 1){
que2.push(que1.front());
que1.pop();
}
ans = que1.front();
que1.pop();
que2.push(ans);
}
else{
while (que2.size() > 1){
que1.push(que2.front());
que2.pop();
}
ans = que2.front();
que2.pop();
que1.push(ans);
}
return ans;
}

// Return whether the stack is empty.
bool empty() {
return que1.empty() && que2.empty();
}
private:
queue<int> q
4000
ue1;
queue<int> que2;
};