两个栈实现一个队列

　　这道题面试的时候经常遇到，特此收录。全部代码来自《剑指offer》。

template <typename T>
class CQueue
{
public:
CQueue(void);
~CQueue(void);

void appendTail(const T &node);
T deleteHead();
private:
stack<T> stack1;
stack<T> stack2;
};

template <typename T>
void CQueue<T>::appendTail(const T &node)
{
stack1.push(node);
}

template <typename T>
T CQueue<T>::deleteHead()
{
if (stack2.size() <= 0)
{
while (stack1.size() > 0)
{
T &data = stack1.top();
stack1.pop();
stack2.push(data);
}
}
if(stack2.size() == 0)
throw new exception("queue is empty");
T head = stack2.top();
stack2.pop();
return head;
}

LeetCode上一道题 Implement Queue using Stacks

solution：

class Queue {
private:
stack<int> stk1, stk2;
public:
// Push element x to the back of queue.
void push(int x) {
stk1.push(x);
}

// Removes the element from in front of queue.
void pop(void) {
peek();
stk2.pop();
}

// Get the front element.
int peek(void) {
if (stk2.empty())
{
while (stk1.size() > 0)
{
stk2.push(stk1.top());
stk1.pop();
}
}
return stk2.top();
}

// Return whether the queue is empty.
bool empty(void) {
return stk1.empty() && stk2.empty();
}
};

PS:

　　两个队列实现一个栈/article/4631369.html

LeetCode上一道题： Implement Stack using Queues

solution： //使用一个队列实现栈

class Stack {
private:
queue<int> que;
public:
// Push element x onto stack.
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();
}
};