此类的目的是计算用户输入的文本中每个n字符单词的出现次数。我认为在methodsNumber方法中有一些错误，因为我为每个n字符单词获得228。错误在哪里？
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
public class TempTest {
public static void main(String[] args) {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the integer n");
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
char[] array = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', ' ' };
StringBuffer[] table = createTotalTable(n, array);
System.out.println("Enter the text ");
StringBuffer text = new StringBuffer("");
try {
text = new StringBuffer(in.readLine());
} catch (IOException a) {
System.out.println("Input-Output problem");
}
StringBuffer text_formatted = format(text.toString());
System.out.println("The formatted text is \n" + text_formatted.toString() + "\n");
System.out.println("Now we print all the n-character words in alphabetic order. Press enter to proceed. ");
try {
in.readLine();
} catch (IOException b) {
System.out.println("Input-Output problem");
}
for (StringBuffer word : table)
System.out.println(word.toString());
int[] occurrenceTable = createList(text_formatted, table, n);
System.out.println("Now we print all the n-words contained in the text with the number of occurrences. Press enter to proceed.");
try {
in.readLine();
} catch (IOException c) {
System.out.println("Input-Output problem");
}
for (int u = 0; u < pow(27, n); u++)
System.out.println(table[u].toString() + ", " + occurrenceTable[u]);
}
public static StringBuffer[] createTotalTable(int n, char[] a) { // this method create an array containing all the n-words in alphabetic order
StringBuffer[] table = new StringBuffer[pow(27, n)];
for (int w = 0; w < pow(27, n); w++)
table[w] = new StringBuffer("");
for (int h = 1; h <= n; h++) {
for (int u = 0; u < pow(27, h - 1); u++) {
for (int j = 0; j < 27; j++) {
for (int x = pow(27, n - h + 1) * u + pow(27, n - h) * j; x < pow(27, n - h + 1) * u + pow(27, n - h) * (j + 1); x++)
table[x] = table[x].append(a[j]);
}
}
}
return table;
}
public static int pow(int a, int b) { // the method Math.pow modified
int tot = 1;
for (int i = 0; i < b; i++)
tot = a * tot;
return tot;
}
public static int occurrenceNumber(StringBuffer testo, StringBuffer parola, int n) { // this method is aimed to calculate the number of occurrences of a
// word of length n in a text
int tot = 0;
if (n > testo.length())
System.out.println("The integer is bigger than the text's length ");
else {
for (int i = 0; i <= testo.length() - n; i++) {
if (testo.substring(i, i + n) == parola.toString())
tot += 1;
}
}
return tot;
}
public static int[] createList(StringBuffer str, StringBuffer[] tabella, int n) { // this method is aimed to create an array containing for every position
// the number of occurrences of the corresponding word in the text
int[] occurrenceTable = new int[pow(27, n)];
for (int i = 0; i < pow(27, n); i++)
occurrenceTable[i] = occurrenceNumber(str, tabella[i], n);
return occurrenceTable;
}
public static StringBuffer format(String s) { // this method is aimed to
// eliminate from the text all non-alphabetic characters and multiple spaces
s = s.toLowerCase();
StringBuffer b = new StringBuffer();
int m = s.length();
int conta_spazi = 0;
StringBuffer h = new StringBuffer(s);
for (int i = 0; i < m; i++) {
switch (h.charAt(i)) {
case 'a':
break;
case 'A':
break;
case 'b':
break;
case 'B':
break;
case 'c':
break;
case 'C':
break;
case 'd':
break;
case 'D':
break;
case 'e':
break;
case 'E':
break;
case 'f':
break;
case 'F':
break;
case 'g':
break;
case 'G':
break;
case 'h':
break;
case 'H':
break;
case 'i':
break;
case 'I':
break;
case 'j':
break;
case 'J':
break;
case 'k':
break;
case 'K':
break;
case 'l':
break;
case 'L':
break;
case 'm':
break;
case 'M':
break;
case 'n':
break;
case 'N':
break;
case 'o':
break;
case 'O':
break;
case 'p':
break;
case 'P':
break;
case 'q':
break;
case 'Q':
break;
case 'r':
break;
case 'R':
break;
case 's':
break;
case 'S':
break;
case 't':
break;
case 'T':
break;
case 'u':
break;
case 'U':
break;
case 'v':
break;
case 'V':
break;
case 'w':
break;
case 'W':
break;
case 'x':
break;
case 'X':
break;
case 'y':
break;
case 'Y':
break;
case 'z':
break;
case 'Z':
break;
default:
h.setCharAt(i, ' ');
}
}
for (int i = 0; i < m; i++) {
if (h.charAt(i) == ' ')
conta_spazi++;
else
conta_spazi = 0;
if (conta_spazi <= 1)
b = b.append(h.charAt(i));
}
return b;
}
}
答案 0 :(得分：3)
if (testo.substring(i, i + n) == parola.toString())
tot += 1;
请勿使用==运算符来比较字符串。使用equals方法。
if (testo.substring(i, i + n).equals(parola.toString()))
tot += 1;
例如，请参阅Java String.equals versus ==或How do I compare strings in Java?
答案 1 :(得分：2)
代码是孤独的，不是很清楚，但如果你有一个字符串包含你可以做的全部事情：
string str = GetTheWholeString();// get the string
int n = 5;// we are searching for 5-lettersWords
String[] words = str.split(' ');
Hashtable wordsAccur = new Hashtable();
for(String currWord : words)
{
if(currWord.length() == n)
{
if(wordsAccur.containsKey(currWord))
{
wordsAccur.put(currWord, (int)wordsAccur.get(currWord) + 1);
}
else
{
wordsAccur.put(currWord, 1);
}
}
}
你可以打印它们，或者用它们做你需要的东西：
Enumeration names;
names = balance.keys();
while(names.hasMoreElements()) {
string curr = (String) names.nextElement();
System.out.println(curr + ": " +
wordsAccur.get(curr));
}