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操作的時間復雜度都是O(1)，且設計的棧類型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.由兩個棧組成的隊列
編寫一個類，用兩個棧實現隊列，支持隊列的基本操作（add、poll、peek）
思路：一個棧正常入棧，之后倒入另一個輔助棧，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.由兩個隊列組成的棧
編寫一個類，用兩個隊列實現棧，支持棧的基本操作（push、poll、peek）

public static class Queue<T>{private LinkedList<T> queue = new LinkedList<T>();//LinkedList標準的雙端隊列，ArrayList是動態數組，在刷題階段時使用鏈表時用LinkedListpublic 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;}}

總結

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