如何在一个字符串中找到一个字符出现的次数?
例如:The quick brown fox jumped over the lazy dog。
以下是一些示例输出:
'a' = 1
'o' = 4
'space' = 8
'.' = 1
如何在一个字符串中找到一个字符出现的次数?
例如:The quick brown fox jumped over the lazy dog。
以下是一些示例输出:
'a' = 1
'o' = 4
'space' = 8
'.' = 1
如果你想处理的是字符串 s
,那么你可以使用以下代码:
Map<Character,Integer> map = new HashMap<Character,Integer>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (map.containsKey(c)) {
int cnt = map.get(c);
map.put(c, ++cnt);
} else {
map.put(c, 1);
}
}
请注意,它会计算所有字符,而不仅仅是字母。
Map
没有 contains
方法。 - Rohit JainJava 8的方式:
"The quick brown fox jumped over the lazy dog."
.chars()
.mapToObj(i -> (char) i)
.collect(Collectors.groupingBy(Object::toString, Collectors.counting()));
void Findrepeter(){
String s="mmababctamantlslmag";
int distinct = 0 ;
for (int i = 0; i < s.length(); i++) {
for (int j = 0; j < s.length(); j++) {
if(s.charAt(i)==s.charAt(j))
{
distinct++;
}
}
System.out.println(s.charAt(i)+"--"+distinct);
String d=String.valueOf(s.charAt(i)).trim();
s=s.replaceAll(d,"");
distinct = 0;
}
}
Map<String, Integer>
替代。另外,您的方法名称似乎有一个拼写错误。 - xlmimport java.io.*;
public class CountChar
{
public static void main(String[] args) throws IOException
{
String ch;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter the Statement:");
ch=br.readLine();
int count=0,len=0;
do
{
try
{
char name[]=ch.toCharArray();
len=name.length;
count=0;
for(int j=0;j<len;j++)
{
if((name[0]==name[j])&&((name[0]>=65&&name[0]<=91)||(name[0]>=97&&name[0]<=123)))
count++;
}
if(count!=0)
System.out.println(name[0]+" "+count+" Times");
ch=ch.replace(""+name[0],"");
}
catch(Exception ex){}
}
while(len!=1);
}
}
输出
Enter the Statement:asdf23123sfsdf
a 1 Times
s 3 Times
d 2 Times
f 3 Times
Map
来存储计数。这将是一个Map<Character, Integer>
。Character#isAlphabetic
方法进行检查。如果是alphabet
,则增加Map
中的计数。如果字符尚未在Map
中,则将其添加为计数1
。Character.isAlphabetic
方法是Java 7
中的新方法。如果您使用较旧版本,则应使用Character#isLetter
。 String str = "asdfasdfafk asd234asda";
Map<Character, Integer> charMap = new HashMap<Character, Integer>();
char[] arr = str.toCharArray();
for (char value: arr) {
if (Character.isAlphabetic(value)) {
if (charMap.containsKey(value)) {
charMap.put(value, charMap.get(value) + 1);
} else {
charMap.put(value, 1);
}
}
}
System.out.println(charMap);
输出: -
{f=3, d=4, s=4, a=6, k=1}
Character
中没有isAlphabetic()
=) 至少在Java 6中是这样的。可能应该使用isLetter()
。 - Andrew Logvinovpublic class a36 {
public static void main(String[] args) {
String a = "Gini Rani";
fix(a);
}//main
public static void fix(String a ){
Map<Character ,Integer> map = new HashMap<>();
for (int i = 0; i <a.length() ; i++ ) {
char ch = a.charAt(i);
map.put(ch , map.getOrDefault(ch,0) +1 );
}//for
List<Character> list = new ArrayList<>();
Set<Map.Entry<Character ,Integer> > entrySet = map.entrySet();
for ( Map.Entry<Character ,Integer> entry : entrySet) {
list.add( entry.getKey() );
System.out.printf( " %s : %d %n" , entry.getKey(), entry.getValue() );
}//for
System.out.println("Duplicate elements => " + list);
}//fix
}
public class a37 {
public static void main(String[] args) {
String aa = "Protijayi Gini";
String[] stringarray = aa.split("");
Map<String , Long> map = Arrays.stream(stringarray)
.collect(Collectors.groupingBy(c -> c , Collectors.counting()));
map.forEach( (k, v) -> System.out.println(k + " : "+ v) );
}
}
public class StringExample {
public static void main(String[] args) {
String str = "abcdabghplhhnfl".toLowerCase();
// create a integer array for 26 alphabets.
// where index 0,1,2.. will be the container for frequency of a,b,c...
Integer[] ar = new Integer[26];
// fill the integer array with character frequency.
for(int i=0;i<str.length();i++) {
int j = str.charAt(i) -'a';
if(ar[j]==null) {
ar[j]= 1;
}else {
ar[j]+= 1;
}
}
// print only those alphabets having frequency greater then 1.
for(int i=0;i<ar.length;i++) {
if(ar[i]!=null && ar[i]>1) {
char c = (char) (97+i);
System.out.println("'"+c+"' comes "+ar[i]+" times.");
}
}
}
}
输出:
'a' comes 2 times.
'b' comes 2 times.
'h' comes 3 times.
'l' comes 2 times.
这是一种不使用任何集合的实现方式,复杂度为n级别。虽然已经接受的解决方案也很好,而且也不使用集合,但似乎它没有处理特殊字符的情况。
import java.util.Arrays;
public class DuplicateCharactersInString {
public static void main(String[] args) {
String string = "check duplicate charcters in string";
string = string.toLowerCase();
char[] charAr = string.toCharArray();
Arrays.sort(charAr);
for (int i = 1; i < charAr.length;) {
int count = recursiveMethod(charAr, i, 1);
if (count > 1) {
System.out.println("'" + charAr[i] + "' comes " + count + " times");
i = i + count;
} else
i++;
}
}
public static int recursiveMethod(char[] charAr, int i, int count) {
if (ifEquals(charAr[i - 1], charAr[i])) {
count = count + recursiveMethod(charAr, ++i, count);
}
return count;
}
public static boolean ifEquals(char a, char b) {
return a == b;
}
}
输出结果:
' ' comes 4 times
'a' comes 2 times
'c' comes 5 times
'e' comes 3 times
'h' comes 2 times
'i' comes 3 times
'n' comes 2 times
'r' comes 3 times
's' comes 2 times
't' comes 3 times
public class dublicate
{
public static void main(String...a)
{
System.out.print("Enter the String");
Scanner sc=new Scanner(System.in);
String st=sc.nextLine();
int [] ar=new int[256];
for(int i=0;i<st.length();i++)
{
ar[st.charAt(i)]=ar[st.charAt(i)]+1;
}
for(int i=0;i<256;i++)
{
char ch=(char)i;
if(ar[i]>0)
{
if(ar[i]==1)
{
System.out.print(ch);
}
else
{
System.out.print(ch+""+ar[i]);
}
}
}
}
}
使用Google Guava的Multiset<String>
。
Multiset<String> wordsMultiset = HashMultiset.create();
wordsMultiset.addAll(words);
for(Multiset.Entry<E> entry:wordsMultiset.entrySet()){
System.out.println(entry.getElement()+" - "+entry.getCount());
}