近來在啃資料結構與演算法，歡迎交流。

1.3.14：用可變長度的陣列實現佇列（FIFO）

• 資料結構ResizingArrayQueueOfStrings 的實現：

public class ResizingArrayQueueOfStrings {
	String[] arr;
	int N; 

	public ResizingArrayQueueOfStrings(int cap) {
		arr = new String[cap];
	}

	public void queue(String str) {
		if (N + 1 == arr.length) 
			resize(arr.length * 2);
		for (int i = N; i >= 0; i--) {
			arr[i + 1] = arr[i];
		}
		arr[0] = str;
		N++;
	}

	public String dequeue() {
		String s = arr[--N];
		if (N > 0 && N == arr.length / 4)
			resize(arr.length / 2);
		return s;
	}

	private void resize(int max) {
		String[] arrnew = new String[max];
		for (int i = 0; i <= N; i++) {
			arrnew[i] = arr[i];
		}
		arr = arrnew;
	
	}
	
	public boolean isEmpty(){
		return N == 0;
	}
	public int size(){
		return N;
	}
}
 

• client code（書上似乎稱作：用例程式碼，翻譯有點拗口喲）

public class Ex1_3_14_Client {

	public static void main(String[] args) {
		ResizingArrayQueueOfStrings qos = new ResizingArrayQueueOfStrings(5);
		qos.queue("Baba ");
		qos.queue("black ");
		System.out.println("size1:"+qos.size());
		qos.queue("sheep ");
		qos.queue("have ");
		System.out.println("size2:"+qos.size());
		qos.queue("you ");
		qos.queue("any ");
		qos.queue("wood ");
		qos.queue("?");
		System.out.println("size3:"+qos.size());
		
		String s = "";
		while (!qos.isEmpty()) {
			s += qos.dequeue();			
		}
		System.out.println(s);		
	}
}
 

• 測試結果：

size1:2
size2:4
size3:8
Baba black sheep have you any wood ?

Baba black sheep have you any wood ?不知道有多少人知道這首朗朗上口的英文兒歌呢~

1.3.19：在以連結串列為基礎的棧中新增一個方法，使之可以進行頭尾“壓棧”、頭尾“彈棧”並返回彈出的元素：

• 資料結構MyLinkedlist實現：

public class MyLinkedlist<Item> {
	private int N;
	private Node first;
	private Node last;
	
	
	private class Node{
		Node next;
		Item item;
	}
	
	public void push(Item item){
		Node oldfirst = first;
		first = new Node();
		first.item = item;
		if (N == 0){
			first.next = null;
			last = first;
		}
		else
			first.next = oldfirst;
		N++;		
	}
		
	public void pushBottom(Item item) {
		Node oldlast = last;
		last = new Node();
		last.item = item;
		last.next = null;
		if (N == 0)
			first = last;
		else 
			oldlast.next = last;
		N++;
	}
	
	public Item pop(){
		Item item = first.item;
		first = first.next;
		N--;
		return item;
	}
	
	public Item popBottom(){
		Item item  = last.item;
		Node temp = first;
		for (int i = 1; i < N-1; i++) {
			temp = temp.next;       			// 第n-1個結點 
		}
		last = temp;
		last.next = null;
		N--;
		return item;
	}
	
	public int size(){
		return N;
	}
}
 

• client code：

public class Ex1_3_19 {

	public static void main(String[] args) {
		MyLinkedlist<String> mll = new MyLinkedlist<>();
		mll.push("wood ");
		mll.push("any ");
		mll.push("you ");
		mll.push("have ");
		mll.pushBottom("? ");
		
		String str = "";
		int len = mll.size();
		for (int i = 0; i < len; i++) {
			str += mll.popBottom();
//              str += mll.pop();
		}
		System.out.println(str);
	}
}

• 測試結果：

popBottom:
? wood any you have 

pop:
have you any wood ?