Implement Stack using Queues - LeetCode 225

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).

Update (2015-06-11):

The class name of the Java function had been updated to MyStack instead of Stack.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and all test cases.

Hide Tags Data Structure

分析：

由于栈是先进后出，队列是先进先出，两者出栈顺序是相反的，因此，因此两个队列来实现一个栈，一个作为中转队列，一个是目标队列，二者不断交换角色。过程描述如下：

在用队列实现栈的出栈操作时，需要删除队队尾的元素，那么得将目标队列的非队尾元素出队列存放到中转队列中，然后出队原队尾元素。此时目标队列和中转队列角色互换。

对于压栈操作，直接将元素放入目标队列（非空）队尾，若两个队列都为空，那么随便选择其中一个作为目标的队列即可。

对于判空，如过两个队列都为空，那么返回true，否则返回false。

对于top操作，直接返回非空队列的队尾元素即可。

思路还是挺简单的，以下是C++实现代码：

/**///////////////////////////0ms*/
class Stack {
public:
queue<int> q1,q2;

// Push element x onto stack.
void push(int x) {
if(q1.empty() && q2.empty() || !q1.empty())
q1.push(x);
else
q2.push(x);
}

// Removes the element on top of the stack.
void pop() {
int top = 0;
if(!Stack::empty()){
if(!q1.empty()){
int n = q1.size()-1;
while(n--)
{
q2.push(q1.front());
q1.pop();
}
q1.pop();
}
else{
int n = q2.size()-1;
while(n--)
{
q2.push(q2.front());
q2.pop();
}
q2.pop();
}
}
}

// Get the top element.
int top(){
if(!Stack::empty()){
if(!q1.empty())
return q1.back();
else
return q2.back();
}
}

// Return whether the stack is empty.
bool empty() {
if(q1.empty() && q2.empty())
return true;
return false;
}
};