我非常接近完成，但不能完全弄清楚如何将所有东西捆绑在一起。我有单独的方法负责他们的特定任务，但我觉得我的凝聚力非常糟糕。不确定它们如何结合在一起以及需要在主要部分中调用什么。这里的目标是从命令行中读取文本文件，并按字典顺序列出故事中的单词。
% java Ten < story.txt
Word Occurs
==== ======
a 21
animal 3
.
.
.
zebra 1
%
到目前为止，这是我的代码：
import java.util.Scanner;
public class Ten2
{
public static void main(String [] arg)
{
Scanner input = new Scanner(System.in);
String word;
List sortedList = new List();
word = nextWord(input);
while (word!=null) {
sortedList.insert(word);
word = nextWord(input);
}
sortedList.printWords();
}
private static String nextWord(Scanner input)
{
// return null if there are no more words
if (input.hasNext() == false )
return null;
// take next element, convert to lowercase, store in s
else {
String s = input.next().toLowerCase() ;
// empty string
String token = "";
// loop through s and concatonate chars onto token
for (int i =0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i)) == true)
token = token + s.charAt(i);
else if (s.charAt(i) == '\'' )
token = token + s.charAt(i);
else if (s.charAt(i) == '-')
token = token + s.charAt(i);
}
return token;
}
}
}
class List
{
/*
* Internally, the list of strings is represented by a linked chain
* of nodes belonging to the class ListNode. The strings are stored
* in lexicographical order.
*/
private static class ListNode
{
// instance variables for ListNode objects
public String word;
public ListNode next;
public int count;
// Listnode constructor
public ListNode(String w, ListNode nxt)
{
word = w; // token from nextWord()?
next = nxt; // link to next ListNode
count = 1; // number of word occurences
}
}
// instance variables for List object
private ListNode first;
private int numWords;
// constructor postcondition: creates new Listnode storing object
public List()
{
first = null; // pointer to ListNode?
numWords = 0; // counter for nodes in list
}
// Insert a specified word into the list, keeping the list
// in lexicographical order.
public void insert(String word)
{
// check if first is null
if (first == null) {
ListNode newNode;
newNode = addNode(word, null);
first = newNode;
}
// else if (first is not null) check if word matches first word in List
else if (word.equals(first.word)) {
// increase count
first.count++;
}
// else (first is not null && doesn't match first word)
else {
ListNode newNode;
ListNode current;
current = first;
ListNode previous;
previous = null;
int cmp = word.compareTo(current.word);
/*
* Fist two cases of empty list and word already existing
* handled in above if and else statements, now by using compareTo()
* method between the words, the insertion postion can be determined.
* Links between ListNode variables current and previous need to be
* modified in order to maintain the list
*/
// loop as long as value comparing to is positive
// when compareTo() returns positive this means the "word" parameter is greater than the word in the list
while ((cmp >0) && (current.next != null)) {
previous = current;
current = current.next;
cmp = word.compareTo(current.word);
}
// insert after current at end of list
if ((cmp >0 && current.next == null)) {
newNode = addNode(word, null);
current.next = newNode;
}
// increments count when word already exists
else if (cmp==0) {
current.count++;
}
// else (cmp < 0) we insert BEFORE current
else {
newNode = addNode(word, current);
// first node in list comes after new word
if (previous == null) {
first = newNode;
}
else {
// inserting new word in middle of list
previous.next = newNode;
}
}
}
}
// method to add new ListNode and increase counter
private ListNode addNode(String word, ListNode next)
{
ListNode newNode = new ListNode(word, next);
numWords++;
return newNode;
}
// Returns a string array that contains all the words in the list.
public String[] getWords()
{
String[] Words = new String[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Words[i] = current.word;
current = current.next;
i++;
}
return Words;
}
// Returns an int array that contains the number of times
// each word occurs in the list.
public int[] getNumbers()
{
int[] Numbers = new int[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Numbers[i] = current.count;
current = current.next;
i++;
}
return Numbers;
}
// Outputs the string array and int array containing all the
// words in the list and the number of times each occurs.
public void printWords()
{
int[] Numbers = getNumbers();
String[] Words = getWords();
System.out.println("Word \t \t Occurs");
System.out.println("==== \t \t ======");
for (int i =0; i < numWords; i++) {
System.out.println(Words[i] + " \t " + Numbers[i]);
}
}
}
答案 0 :(得分：3)
好吧，我首先要定义你希望你的程序做什么，你已经完成了：
此处的目标是从命令行中读取文本文件，并按字典顺序列出故事中的单词。
你的主要功能是几乎。基本上，你需要的是一个将它绑在一起的循环：
public static void main(String [] arg)
{
// print out your initial information first (i.e. your column headers)
// ...
List sortedList = new List();
String word = nextWord();
// now, the question is... what is the end condition of the loop?
// probably that there aren't any more words, so word in this case
// will be null
while (word != null)
{
sortedList.insert(word);
word = nextWord();
}
// now, your list takes care of the sorting, so let's just print the list
sortedList.printWords();
}
我认为这就是它的全部。通常情况下，我不喜欢将解决方案发布到家庭作业问题上，但在这种情况下，由于你已经拥有了所有的代码，而你只是需要一点推动才能将你推向正确的方向，我认为这很好。
我注意到有些事情与您的
不一致您的列表构造函数具有'void'返回类型 - 构造函数上应该没有返回类型：
public List() //make an empty list
{
first = null;
numWords = 0;
}
此方法中的'else'语句不需要：
public static String nextWord()
{
if ( keyboard.hasNext() == false )
return null;
else {
String start = keyboard.next().toLowerCase() ;
String organized = "";
for (int i =0; i < start.length(); i++) {
if (Character.isLetter(start.charAt(i)) == true)
organized = organized + start.charAt(i);
else if (start.charAt(i) == '\'' )
organized = organized + start.charAt(i);
else if (start.charAt(i) == '-')
organized = organized + start.charAt(i);
}
return organized;
}
}
所以，这应该是：
public static String nextWord()
{
if ( keyboard.hasNext() == false )
return null;
String start = keyboard.next().toLowerCase() ;
String organized = "";
for (int i =0; i < start.length(); i++) {
if (Character.isLetter(start.charAt(i)) == true)
organized = organized + start.charAt(i);
else if (start.charAt(i) == '\'' )
organized = organized + start.charAt(i);
else if (start.charAt(i) == '-')
organized = organized + start.charAt(i);
}
return organized;
}
如果你想使用BufferedReader，这很容易。只需在主方法中进行设置：
if (arg.length > 0)
{
// open our file and read everything into a string buffer
BufferedReader bRead = null;
try {
bRead = new BufferedReader(new FileReader(arg[0]));
} catch(FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
System.exit(0);
}
setupScanner(bRead);
}
然后，创建一个设置扫描仪对象的新方法：
public static void setupScanner(BufferedReader rdr)
{
keyboard = new Scanner(rdr);
}
然后在命令行中传入它（即java ten2 [filename]）
答案 1 :(得分：0)
import java.util.Scanner;
public class Ten2
{
public static void main(String [] arg)
{
Scanner input = new Scanner(System.in);
String word;
List sortedList = new List();
word = nextWord(input);
while (word!=null) {
sortedList.insert(word);
word = nextWord(input);
}
sortedList.printWords();
}
private static String nextWord(Scanner input)
{
// return null if there are no more words
if (input.hasNext() == false )
return null;
// take next element, convert to lowercase, store in s
else {
String s = input.next().toLowerCase() ;
// empty string
String token = "";
// loop through s and concatonate chars onto token
for (int i =0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i)) == true)
token = token + s.charAt(i);
else if (s.charAt(i) == '\'' )
token = token + s.charAt(i);
else if (s.charAt(i) == '-')
token = token + s.charAt(i);
}
return token;
}
}
}
class List
{
/*
* Internally, the list of strings is represented by a linked chain
* of nodes belonging to the class ListNode. The strings are stored
* in lexicographical order.
*/
private static class ListNode
{
// instance variables for ListNode objects
public String word;
public ListNode next;
public int count;
// Listnode constructor
public ListNode(String w, ListNode nxt)
{
word = w; // token from nextWord()?
next = nxt; // link to next ListNode
count = 1; // number of word occurences
}
}
// instance variables for List object
private ListNode first;
private int numWords;
// constructor postcondition: creates new Listnode storing object
public List()
{
first = null; // pointer to ListNode?
numWords = 0; // counter for nodes in list
}
// Insert a specified word into the list, keeping the list
// in lexicographical order.
public void insert(String word)
{
// check if first is null
if (first == null) {
ListNode newNode;
newNode = addNode(word, null);
first = newNode;
}
// else if (first is not null) check if word matches first word in List
else if (word.equals(first.word)) {
// increase count
first.count++;
}
// else (first is not null && doesn't match first word)
else {
ListNode newNode;
ListNode current;
current = first;
ListNode previous;
previous = null;
int cmp = word.compareTo(current.word);
/*
* Fist two cases of empty list and word already existing
* handled in above if and else statements, now by using compareTo()
* method between the words, the insertion postion can be determined.
* Links between ListNode variables current and previous need to be
* modified in order to maintain the list
*/
// loop as long as value comparing to is positive
// when compareTo() returns positive this means the "word" parameter is greater than the word in the list
while ((cmp >0) && (current.next != null)) {
previous = current;
current = current.next;
cmp = word.compareTo(current.word);
}
// insert after current at end of list
if ((cmp >0 && current.next == null)) {
newNode = addNode(word, null);
current.next = newNode;
}
// increments count when word already exists
else if (cmp==0) {
current.count++;
}
// else (cmp < 0) we insert BEFORE current
else {
newNode = addNode(word, current);
// first node in list comes after new word
if (previous == null) {
first = newNode;
}
else {
// inserting new word in middle of list
previous.next = newNode;
}
}
}
}
// method to add new ListNode and increase counter
private ListNode addNode(String word, ListNode next)
{
ListNode newNode = new ListNode(word, next);
numWords++;
return newNode;
}
// Returns a string array that contains all the words in the list.
public String[] getWords()
{
String[] Words = new String[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Words[i] = current.word;
current = current.next;
i++;
}
return Words;
}
// Returns an int array that contains the number of times
// each word occurs in the list.
public int[] getNumbers()
{
int[] Numbers = new int[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Numbers[i] = current.count;
current = current.next;
i++;
}
return Numbers;
}
// Outputs the string array and int array containing all the
// words in the list and the number of times each occurs.
public void printWords()
{
int[] Numbers = getNumbers();
String[] Words = getWords();
System.out.println("Word \t \t Occurs");
System.out.println("==== \t \t ======");
for (int i =0; i < numWords; i++) {
System.out.println(Words[i] + " \t " + Numbers[i]);
}
}
}