我想知道如何编写一个方法来计算java字符串中的单词数,只能使用字符串方法,如charAt,length或substring。
循环和if语句没问题!
我非常感谢能得到的任何帮助!谢谢!
答案 0 :(得分:67)
即使有多个空格和前导和/或尾随空格和空行,这也可以工作:
String trim = s.trim();
if (trim.isEmpty())
return 0;
return trim.split("\\s+").length; // separate string around spaces
希望有所帮助。有关split here.
的更多信息答案 1 :(得分:24)
public static int countWords(String s){
int wordCount = 0;
boolean word = false;
int endOfLine = s.length() - 1;
for (int i = 0; i < s.length(); i++) {
// if the char is a letter, word = true.
if (Character.isLetter(s.charAt(i)) && i != endOfLine) {
word = true;
// if char isn't a letter and there have been letters before,
// counter goes up.
} else if (!Character.isLetter(s.charAt(i)) && word) {
wordCount++;
word = false;
// last word of String; if it doesn't end with a non letter, it
// wouldn't count without this.
} else if (Character.isLetter(s.charAt(i)) && i == endOfLine) {
wordCount++;
}
}
return wordCount;
}
答案 2 :(得分:11)
您好我刚刚想到了StringTokenizer:
String words = "word word2 word3 word4";
StringTokenizer st = new Tokenizer(words);
st.countTokens();
答案 3 :(得分:7)
只需使用,
str.split("\\w+").length ;
答案 4 :(得分:2)
public static int countWords(String str){
if(str == null || str.isEmpty())
return 0;
int count = 0;
for(int e = 0; e < str.length(); e++){
if(str.charAt(e) != ' '){
count++;
while(str.charAt(e) != ' ' && e < str.length()-1){
e++;
}
}
}
return count;
}
答案 5 :(得分:1)
private static int countWordsInSentence(String input) {
int wordCount = 0;
if (input.trim().equals("")) {
return wordCount;
}
else {
wordCount = 1;
}
for (int i = 0; i < input.length(); i++) {
char ch = input.charAt(i);
String str = new String("" + ch);
if (i+1 != input.length() && str.equals(" ") && !(""+ input.charAt(i+1)).equals(" ")) {
wordCount++;
}
}
return wordCount;
}
答案 6 :(得分:0)
有一个简单的解决方案您可以尝试此代码
String s = "hju vg jhdgsf dh gg g g g ";
String[] words = s.trim().split("\\s+");
System.out.println("count is = "+(words.length));
答案 7 :(得分:0)
lambda,其中省去了计数单词的拆分和存储
,并且仅进行了计数
String text = "counting w/o apostrophe's problems or consecutive spaces";
int count = text.codePoints().boxed().collect(
Collector.of(
() -> new int[] {0, 0},
(a, c) -> {
if( ".,; \t".indexOf( c ) >= 0 )
a[1] = 0;
else if( a[1]++ == 0 ) a[0]++;
}, (a, b) -> {a[0] += b[0]; return( a );},
a -> a[0] ) );
得到:7
用作状态机,计算从空格字符.,; \t
到单词的转换
答案 8 :(得分:0)
创建变量计数,状态。初始化变量
如果有空间,请保持计数,否则增加计数。
例如:
if (string.charAt(i) == ' ' ) {
state = 0;
} else if (state == 0) {
state = 1;
count += 1;
答案 9 :(得分:0)
我把它放在一起。 wordCount()方法中的增量器对我来说有点不合适,但它有效。
import java.util.*;
public class WordCounter {
private String word;
private int numWords;
public int wordCount(String wrd) {
StringTokenizer token = new StringTokenizer(wrd, " ");
word = token.nextToken();
numWords = token.countTokens();
numWords++;
return numWords;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String userWord;
WordCounter wc = new WordCounter();
System.out.println("Enter a sentence.");
userWord = input.nextLine();
wc.wordCount(userWord);
System.out.println("You sentence was " + wc.numWords + " words long.");
}
}
答案 10 :(得分:0)
以所选答案为出发点,以下内容涉及一些英语语言问题,包括带连字符的单词,占有和缩写的撇号,数字以及UTF-16以外的任何字符:
public static int countWords(final String s) {
int wordCount = 0;
boolean word = false;
final int endOfLine = s.length() - 1;
for (int i = 0; i < s.length(); i++) {
// if the char is a letter, word = true.
if (isWordCharacter(s, i) && i != endOfLine) {
word = true;
// if char isn't a letter and there have been letters before,
// counter goes up.
} else if (!isWordCharacter(s, i) && word) {
wordCount++;
word = false;
// last word of String; if it doesn't end with a non letter, it
// wouldn't count without this.
} else if (isWordCharacter(s, i) && i == endOfLine) {
wordCount++;
}
}
return wordCount;
}
private static boolean isWordCharacter(final String s, final int i) {
final char ch = s.charAt(i);
return Character.isLetterOrDigit(ch)
|| ch == '\''
|| Character.getType(ch) == Character.DASH_PUNCTUATION
|| Character.isSurrogate(ch);
}
答案 11 :(得分:0)
字符串短语normaly具有以空格分隔的单词。那么你可以使用空格分隔短语作为分隔字符并按如下方式计算。
import java.util.HashMap;
import java.util.Map;
public class WordCountMethod {
public static void main (String [] args){
Map<String, Integer>m = new HashMap<String, Integer>();
String phrase = "hello my name is John I repeat John";
String [] array = phrase.split(" ");
for(int i =0; i < array.length; i++){
String word_i = array[i];
Integer ci = m.get(word_i);
if(ci == null){
m.put(word_i, 1);
}
else m.put(word_i, ci+1);
}
for(String s : m.keySet()){
System.out.println(s+" repeats "+m.get(s));
}
}
}
答案 12 :(得分:0)
我是stackoverflow的新手,但我希望我的代码有所帮助:
private int numOfWordsInLineCounter(String line){
int words = 0;
for(int i = 1 ; i<line.length();i++){
Character ch = line.charAt(i-1);
Character bch = line.charAt(i);
if(Character.isLetterOrDigit(ch) == true && Character.isLetterOrDigit(bch)== false ) words++;
if(i == line.length()-1 && Character.isLetterOrDigit(bch))words++;
}
return words;
}
答案 13 :(得分:0)
我对该计划的想法是:
package text;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class CoutingWords {
public static void main(String[] args) throws IOException {
String str;
int cWords = 1;
char ch;
BufferedReader buffor = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter text: ");
str = buffor.readLine();
for(int i =0; i<str.length(); i++){
ch = str.charAt(i);
if(Character.isWhitespace(ch)){ cWords++; }
}
System.out.println("There are " + (int)cWords +" words.");
}
}
答案 14 :(得分:0)
public static int countWords(String input) {
int wordCount = 0;
boolean isBlankSet = false;
input = input.trim();
for (int j = 0; j < input.length(); j++) {
if (input.charAt(j) == ' ')
isBlankSet = true;
else {
if (isBlankSet) {
wordCount++;
isBlankSet = false;
}
}
}
return wordCount + 1;
}
答案 15 :(得分:0)
使用
myString.split("\\s+");
这样可行。
答案 16 :(得分:0)
import java.util。; import java.io。;
public class Main {
public static void main(String[] args) {
File f=new File("src/MyFrame.java");
String value=null;
int i=0;
int j=0;
int k=0;
try {
Scanner in =new Scanner(f);
while(in.hasNextLine())
{
String a=in.nextLine();
k++;
char chars[]=a.toCharArray();
i +=chars.length;
}
in.close();
Scanner in2=new Scanner(f);
while(in2.hasNext())
{
String b=in2.next();
System.out.println(b);
j++;
}
in2.close();
System.out.println("the number of chars is :"+i);
System.out.println("the number of words is :"+j);
System.out.println("the number of lines is :"+k);
}
catch (Exception e) {
e.printStackTrace();
}
}
}
答案 17 :(得分:0)
计算字符串中的单词:
这也可能有用 - >
package data.structure.test;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class CountWords {
public static void main(String[] args) throws IOException {
// Couting number of words in a string
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter Your String");
String input = br.readLine();
char[] arr = input.toCharArray();
int i = 0;
boolean notCounted = true;
int counter = 0;
while (i < arr.length) {
if (arr[i] != ' ') {
if (notCounted) {
notCounted = false;
counter++;
}
} else {
notCounted = true;
}
i++;
}
System.out.println("words in the string are : " + counter);
}
}
答案 18 :(得分:0)
import com.google.common.base.Optional;
import com.google.common.base.Splitter;
import com.google.common.collect.HashMultiset;
import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Multiset;
String str="Simple Java Word Count count Count Program";
Iterable<String> words = Splitter.on(" ").trimResults().split(str);
//google word counter
Multiset<String> wordsMultiset = HashMultiset.create();
for (String string : words) {
wordsMultiset.add(string.toLowerCase());
}
Set<String> result = wordsMultiset.elementSet();
for (String string : result) {
System.out.println(string+" X "+wordsMultiset.count(string));
}
add at the pom.xml
<dependency>
<groupId>com.google.guava</groupId>
<artifactId>guava</artifactId>
<version>r09</version>
</dependency>
答案 19 :(得分:0)
public class TestStringCount {
public static void main(String[] args) {
int count=0;
boolean word= false;
String str = "how ma ny wo rds are th ere in th is sente nce";
char[] ch = str.toCharArray();
for(int i =0;i<ch.length;i++){
if(!(ch[i]==' ')){
for(int j=i;j<ch.length;j++,i++){
if(!(ch[j]==' ')){
word= true;
if(j==ch.length-1){
count++;
}
continue;
}
else{
if(word){
count++;
}
word = false;
}
}
}
else{
continue;
}
}
System.out.println("there are "+(count)+" words");
}
}
答案 20 :(得分:0)
Algo in O(N)
count : 0;
if(str[0] == validChar ) :
count++;
else :
for i = 1 ; i < sizeOf(str) ; i++ :
if(str[i] == validChar AND str[i-1] != validChar)
count++;
end if;
end for;
end if;
return count;
答案 21 :(得分:-1)
if(str.isEmpty() || str.trim().length() == 0){
return 0;
}
return (str.trim().split("\\s+").length);
答案 22 :(得分:-1)
String a = "Some String";
int count = 0;
for (int i = 0; i < a.length(); i++) {
if (Character.isWhitespace(a.charAt(i))) {
count++;
}
}
System.out.println(count+1);
它将计算空格。但是,如果我们在count中加1,我们可以获得准确的单词。