leetcode 232: Implement Queue using Stacks
问题描述：
Implement the following operations of a queue using stacks.
push(x) – Push element x to the back of queue.
pop() – Removes the element from in front of queue.
peek() – Get the front element.
empty() – Return whether the queue is empty.
Notes:
You must use only standard operations of a stack – which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
思路：
就是用一个 temp 的 stack 来辅助操作。有两种做法，一种是在长期存储的 stack 存放的是队列顺序，那么 push 操作就需要辅助，pop 和 peek 都不需要；第二种是反过来，长期存储的 stack 存放的仍是栈顺序，那么 push 操作就不需要辅助，pop 和 peek 都需要。
代码：
#include<iostream>
#include<vector>
#include<map>
#include <string>
#include<stack>
using namespace std;
class Queue {
public:stack<int> queue_by_stack;stack<int> temp;// Push element x to the back of queue.void push(int x) {while (!queue_by_stack.empty()) //reverse in the temp stack {temp.push(queue_by_stack.top());queue_by_stack.pop();}queue_by_stack.push(x);while (!temp.empty()) //reverse back to the queue stack {queue_by_stack.push(temp.top());temp.pop();}}// Removes the element from in front of queue.void pop(void) {if (!queue_by_stack.empty())queue_by_stack.pop();}// Get the front element.int peek(void) {if (!queue_by_stack.empty()){return queue_by_stack.top();}elsereturn -999;}// Return whether the queue is empty.bool empty(void) {return queue_by_stack.empty();}
};int main() //for test
{Queue test;test.push(1);test.push(2);cout << "Should be 1: " << test.peek() << endl;test.pop();if (test.empty()){cout << "empty now" << endl;}cout << endl << "Should be 2: " << test.peek() << endl;test.pop();if (test.empty()){cout << "empty now" << endl;}return 0;
}