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).

思路：两个队列模拟栈，入栈选不为空的queue入，出栈先把有元素的queue元素转到另一个，留一个，哪一个就是要出栈的

class Stack {
queue<int> q1;
queue<int> q2;
public:
// Push element x onto stack.
void push(int x) {
if(q1.empty())
{
q2.push(x);
}
else
{
q1.push(x);
}
}

// Removes the element on top of the stack.
void pop() {
if(q1.empty())
{
transElement(2);
q2.pop();
}
else
{
transElement(1);
q1.pop();
}
}

// Get the top element.
int top() {
int result;
if(q1.empty())
{
//q2转q1
transElement(2);
result = q2.front();
q1.push(q2.front());
q2.pop();
}
else
{
transElement(1);
result = q1.front();
q2.push(q1.front());
q1.pop();
}
return result;
}

// Return whether the stack is empty.
bool empty() {
return q1.empty() && q2.empty();
}
//1为q1->q2,2为q2->q1
void transElement(int num)
{
if(num == 1)
{
while(q1.size() > 1)
{
q2.push(q1.front());
q1.pop();
}
}
else
{
while(q2.size() > 1)
{
q1.push(q2.front());
q2.pop();
}
}
}

};