1.設計一個有getMin功能的棧
實現一個特殊的棧，在實現棧的基本功能的基礎上，再實現返回棧中最小元素的操作
要求：peek、pop、push、getMin操作的時間複雜度都是O(1)，且設計的棧型別可以使用現成的棧結構
思路：建立兩個棧，一個data棧作為原始棧，一個help棧為小頂棧，跟data同步壓棧，如果新進棧的數比當前help棧頂元素大，則繼續壓入棧頂的元素，否則壓入棧頂，遍歷一遍即可解決
例如：
|  4  |     |  3  |
|  6  |     |  3  |
|  3  |     |  3  |
|  5  |     |  4  |
|  4  |     |  4  |
|_5_|     |_5_|
data       help

 即，此時help棧頂即為data棧中的最小值，同步壓棧

 public static class MyStack1{
 	private Stack<Integer> stackData;
	private Stack<Integer> stackMin;
	
	public MyStack1(){
		this.stackData = new Stack<Integer>();
		this.stackMin = new Stack<Integer>();
	}
	public void push(){
		if(this.stackMin.isEmpty()){
			this.stackMin.push(newNum);
		}else if(newNum <= this.getmin()){
			this.stackMin.push(newNum);
		}
		this.stackData.push(newNum);
	}
	public int pop(){
		if(this.stackData.isEmpty()){
			throw new RuntimeException("Your stack is empty!");
		}
		int value = this.stackData.pop();
		if(value == this.getmin()){
			this.stackMin.pop();
		}
		return value;
	}
	public int geimin(){
		if(this.stackMin.isEmpty()){
			throw new RuntimeException("Your stack is empty!");
		}
		return this.stackMin.peek();
	}
 }
 public static class MyStack2{
 	private Stack<Integer> stackData;
	private Stack<Integer> stackMin;
	
	public MyStack1(){
		this.stackData = new Stack<Integer>();
		this.stackMin = new Stack<Integer>();
	}
	public void push(){
		if(this.stackMin.isEmpty()){
			this.stackMin.push(newNum);
		}else if(newNum < this.getmin()){
			this.stackMin.push(newNum);
		}else{
			int newMin = this.stackMin.peek();
			this.stackMin.push(newMin);
		}
		this.stackData.push(newNum);
	}
	public int pop(){
		if(this.stackData.isEmpty()){
			throw new RuntimeException("Your stack is empty!");
		}
		this.stackMin.pop();
		return this.stackData.pop();
	}
	public int geimin(){
		if(this.stackMin.isEmpty()){
			throw new RuntimeException("Your stack is empty!");
		}
		return this.stackMin.peek();
	}
 } 
 

2.由兩個棧組成的佇列
編寫一個類，用兩個棧實現佇列，支援佇列的基本操作（add、poll、peek）
思路：一個棧正常入棧，之後倒入另一個輔助棧，輔助棧倒出即可實現。特別注意兩點需要控制，倒入help輔助棧前，help必須為空，倒入help棧時，必須將data棧排空

public static class TwoStackQueue{
	public Stack<Integer> stackPush;
	public Stack<Integer> stackPop;
	public TwoStackQueue(){
		stackPush = new Stack<Integer>();
		stackPop = new Stack<Integer>();
	}
	public void add(int pushInt){
		stackPush.push(pushInt);
	}
	public int poll(){
		if(stackPop.empty() && stackPush.empty()){
			throw new RuntimeException("Queue is empty!");
		}else if(stackPop.empty()){
			while(!stackPush.empty()){
				stackPop.push(stackPush.pop());
			}
		}
		return stackPop.pop();
	}
	public int peek(){
		if(stackPop.empty() && stackPush.empty()){
			throw new RuntimeException("Queue is empty!");
		}else if(stackPop.empty()){
			while(!stackPush.empty()){
				stackPop.push(stackPush.pop());
			}
		}
		return stackPop.peek();
	}
}
 

3.3.由兩個佇列組成的棧
編寫一個類，用兩個佇列實現棧，支援棧的基本操作（push、poll、peek）

public static class Queue<T>{
	private LinkedList<T> queue = new LinkedList<T>();
	//LinkedList標準的雙端佇列，ArrayList是動態陣列，在刷題階段時使用連結串列時用LinkedList
	public boolean isEmpty(){
		return queue.isEmpty();
	}
	public int size(){
		return queue.size();
	}
	public void add(T value){
		queue.addFirst(value);
	}
	public T pull(){
		return queue.pullLast();
	}
	public T peek(){
		return queue.peek();
	}
}
public static class Stack<T>{
	private Queue<T> dQueue = new Queue<T>();
	private Queue<T> hQueue = new Queue<T>();
	
	public boolean isEmpty(){
		return dQueue.isEmpty();
	}
	public int size(){
		return dQueue.size();
	}
	public void push(T value){
		dQueue.add(value);
	}
	public T pop(){
		if(dQueue.isEmpty()){
			throw new RuntimeException("Queue is empty!");
		}
		while(dQueue.size()!=1){
			hQueue.add(dQueue.pull());
		}
		T res = dQueue.pull();
		while(!hQueue.isEmpty()){
			dQueue.add(hQueue.pull());
		}
		return res;
	}
	public T peek(){
		if(dQueue.isEmpty()){
			throw new RuntimeException("Queue is empty!");
		}
		while(dQueue.size()!=1){
			hQueue.add(dQueue.pull());
		}
		T res = dQueue.pull();
		hQueue.add(res);
		while(!hQueue.isEmpty()){
			dQueue.add(hQueue.pull());
		}
		return res;
	}
	
}