Leetcode 225:Implement Stack using Queues

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).

Update (2015-06-11):
The class name of the Java function had been updated to MyStack instead of Stack.

方法一：

用两个队列实现一个栈：
1.压入时，选择不为空的一个队列压入；
2.弹出或求栈顶元素时，首先将非空队列中的所有元素（除队尾元素）弹出保存到另一个队列，然后将队尾元素弹出即可
注意求栈顶元素时，记得将队尾元素也保存到另一个队列。

//用两个队列实现一个栈：//1.压入时，选择不为空的一个队列压入；//2.弹出或求栈顶元素时，首先将非空队列中的所有元素（除队尾元素）弹出保存到另一个队列，然后将队尾元素弹出即可//注意求栈顶元素时，记得将队尾元素也保存到另一个队列class Stack {public:// Push element x onto stack.queue<int> que1;queue<int> que2;void push(int x) {if (!que2.empty()){que2.push(x);}else{que1.push(x);}}// Removes the element on top of the stack.void pop() {if (!que2.empty()){while (que2.size()>1){que1.push(que2.front());que2.pop();}que2.pop();}else{while (que1.size()>1){que2.push(que1.front());que1.pop();}que1.pop();}}// Get the top element.int top() {int topElement = 0;if (!que2.empty()){while (que2.size()>1){que1.push(que2.front());que2.pop();}topElement = que2.front();que1.push(que2.front());que2.pop();}else{while (que1.size()>1){que2.push(que1.front());que1.pop();}topElement = que1.front();que2.push(que1.front());que1.pop();}return topElement;}// Return whether the stack is empty.bool empty() {return que1.empty() && que2.empty();}};

方法二：

用一个队列实现一个栈：
只需要在压入的方式稍作变化就行，每次压入一个元素，按照队列先进先出的原则，新压入的元素都在队尾，我们只要将整个队列翻转，使得队尾元素移动至队头即可，这样我们后面弹出和求栈顶元素的操作，就和队列一样处理就行。

//用一个队列实现一个栈class Stack {public:// Push element x onto stack.queue<int> que;void push(int x) {que.push(x);for (int i = 0; i<que.size() - 1; ++i){que.push(que.front());que.pop();}}// Removes the element on top of the stack.void pop() {que.pop();}// Get the top element.int top() {return que.front();}// Return whether the stack is empty.bool empty() {return que.empty();}};