【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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.
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top() -- Get the top element.
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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.
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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.
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You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
//问题：用队列实现栈
/*
方法一：用两个队列
插入 q1借助辅助队q2在队首插入元素
删除 q1队首元素出队
访问 访问q1队首元素
判断 判断q1是否为空
(Two Queues, push - O(n), pop O(1)
*/
#include <queue>
class MyStack
{
private:
queue<int> q1,q2;
public:
/** Initialize your data structure here. */
MyStack()
{
}
/** Push element x onto stack. */
void push(int x)
{
while (!q1.empty()) //q1转移到q2
{
q2.push(q1.front());
q1.pop();
}
q1.push(x);//将元素插入到q1中（处于队首位置）
while (!q2.empty()) //q2转移回q1
{
q1.push(q2.front());
q2.pop();
}
}
/** Removes the element on top of the stack and returns that element. */
int pop()
{
int top_element = q1.front();
q1.pop();
return top_element;
}
/** Get the top element. */
int top()
{
return q1.front();
}
/** Returns whether the stack is empty. */
bool empty()
{
return q1.empty();
}
};
/*
方法二：用两个队列
插入 在queue1中插入
删除 quue1出队，queue2入队，将queue1中除队尾元素外全部转移到queue2中，再删除queue1中元素即可，最后再q2中元素复制过去q1 (主队q1借助辅助队q2将队尾元素删除)
访问 访问queue1队尾元素
判断是否为空 判断queue1是否为空
注意：stack只有top,某次只能访问栈顶元素，而queue有front,back，某次可以访问队首和队尾元素，不过只能删除队首元素
Two Queues, push O(1), pop O(n)
*/
#include <queue>
class MyStack
{
private:
queue<int> q1,q2;
public:
/** Initialize your data structure here. */
MyStack()
{
}
/** Push element x onto stack. */
void push(int x)
{
q1.push(x);
}
/** Removes the element on top of the stack and returns that element. */
int pop()
{
int top_element = top();
while(q1.size() > 1) //为删除队尾元素，将q1元素除队尾元素外转移到q2
{
q2.push(q1.front());
q1.pop();//删除第一个元素
}
q1.pop(); //删除队尾元素
while(!q2.empty()) //将元素再转移到q1
{
q1.push(q2.front());
q2.pop();
}
return top_element;
}
/** Get the top element. */
int top()
{
return q1.back(); //返回队尾元素即为栈顶元素
}
/** Returns whether the stack is empty. */
bool empty()
{
return q1.empty();
}
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* bool param_4 = obj.empty();
*/
//方法三：一个队列
//每次push某值后，将队首元素依次push到后面并pop队首元素，直到该值称为队首新的队首元素
/*
class Stack {
public:
queue<int> que;
// 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();
}
};
*/