如何查找字符串中某个字符的出现次数？
例如：敏捷的棕色狐狸跳过了懒惰的狗。
下面是一些示例输出，
'a' = 1
'o' = 4
'space' = 8
'.' = 1
问问题
289891 次
32 回答
29
您可以使用以下内容，前提是 Strings是您要处理的字符串。
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，它将计算所有字符，而不仅仅是字母。
于 2012-10-29T10:40:24.233 回答
13
Java 8 方式：
"The quick brown fox jumped over the lazy dog."
.chars()
.mapToObj(i -> (char) i)
.collect(Collectors.groupingBy(Object::toString, Collectors.counting()));
于 2016-10-13T12:54:46.970 回答
10
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;
}
}
于 2014-05-11T11:05:01.597 回答
7
import 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
于 2012-10-29T10:39:42.173 回答
6
更好的方法是创建一个Map来存储您的计数。那将是一个Map<Character, Integer>
您需要遍历字符串的每个字符，并检查它是否是字母表。您可以为此使用Character#isAlphabetic方法。如果是alphabet，则增加其在 中的计数Map。如果该字符尚未在 中，Map则将其添加为1.
注意： -Character.isAlphabetic方法是新的Java 7。如果您使用的是旧版本，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}
于 2012-10-29T10:42:50.097 回答
4
如果你的字符串只包含字母，那么你可以使用这样的东西。
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.
于 2018-02-20T15:25:55.253 回答
4
在 a 中查找重复项String：
示例 1：使用 HashMap
public 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
}
示例 2：在 Java 8 中使用 Arrays.stream()
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) );
}
}
于 2018-04-07T07:10:54.480 回答
3
这是不使用任何 Collection 且复杂度为 n 的实现。尽管公认的解决方案足够好并且也不使用 Collection，但它似乎并没有处理特殊字符。
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 ：
' ' 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
于 2018-02-10T16:49:50.653 回答
2
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]);
}
}
}
}
}
于 2017-08-04T19:11:18.280 回答
1
使用谷歌番石榴Multiset<String>。
Multiset<String> wordsMultiset = HashMultiset.create();
wordsMultiset.addAll(words);
for(Multiset.Entry<E> entry:wordsMultiset.entrySet()){
System.out.println(entry.getElement()+" - "+entry.getCount());
}
于 2012-10-29T10:46:03.363 回答
1
import java.util.HashMap;
import java.util.Scanner;
public class HashMapDemo {
public static void main(String[] args) {
//Create HashMap object to Store Element as Key and Value
HashMap<Character,Integer> hm= new HashMap<Character,Integer>();
//Enter Your String From Console
System.out.println("Enter an String:");
//Create Scanner Class Object From Retrive the element from console to our java application
Scanner sc = new Scanner(System.in);
//Store Data in an string format
String s1=sc.nextLine();
//find the length of an string and check that hashmap object contain the character or not by using
//containskey() if that map object contain element only one than count that value as one or if it contain more than one than increment value
for(int i=0;i<s1.length();i++){
if(!hm.containsKey(s1.charAt(i))){
hm.put(s1.charAt(i),(Integer)1);
}//if
else{
hm.put(s1.charAt(i),hm.get(s1.charAt(i))+1);
}//else
}//for
System.out.println("The Charecters are:"+hm);
}//main
}//HashMapDemo
于 2015-09-14T15:59:13.293 回答
1
public static void main(String args[]) {
char Char;
int count;
String a = "Hi my name is Rahul";
a = a.toLowerCase();
for (Char = 'a'; Char <= 'z'; Char++) {
count = 0;
for (int i = 0; i < a.length(); i++) {
if (a.charAt(i) == Char) {
count++;
}
}
System.out.println("Number of occurences of " + Char + " is " + count);
}
}
于 2017-01-24T02:11:14.937 回答
1
public static void main(String[] args) {
String name="AnuvratAnuvra";
char[] arr = name.toCharArray();
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
for(char val:arr){
map.put(val,map.containsKey(val)?map.get(val)+1:1);
}
for (Entry<Character, Integer> entry : map.entrySet()) {
if(entry.getValue()>1){
Character key = entry.getKey();
Object value = entry.getValue();
System.out.println(key + ":"+value);
}
}
}
于 2017-02-13T13:49:07.363 回答
1
A类{
public static void getDuplicates(String S) {
int count = 0;
String t = "";
for (int i = 0; i < S.length() - 1; i++) {
for (int j = i + 1; j < S.length(); j++) {
if (S.charAt(i) == S.charAt(j) && !t.contains(S.charAt(j) + "")) {
t = t + S.charAt(i);
}
}
}
System.out.println(t);
}
}
B类公共B类{
public static void main(String[] args){
A.getDuplicates("mymgsgkkabcdyy");
}
}
于 2017-03-15T13:55:05.200 回答
1
使用Eclipse 集合 CharAdapter和CharBag：
CharBag bag =
Strings.asChars("The quick brown fox jumped over the lazy dog.").toBag();
Assert.assertEquals(1, bag.occurrencesOf('a'));
Assert.assertEquals(4, bag.occurrencesOf('o'));
Assert.assertEquals(8, bag.occurrencesOf(' '));
Assert.assertEquals(1, bag.occurrencesOf('.'));
注意：我是 Eclipse Collections 的提交者
于 2017-05-21T06:12:33.060 回答
1
您还可以通过遍历您的String并使用 aswitch检查每个单独的字符来实现它，并在找到匹配项时添加一个计数器。啊，也许一些代码会更清楚：
主要应用：
public static void main(String[] args) {
String test = "The quick brown fox jumped over the lazy dog.";
int countA = 0, countO = 0, countSpace = 0, countDot = 0;
for (int i = 0; i < test.length(); i++) {
switch (test.charAt(i)) {
case 'a':
case 'A': countA++; break;
case 'o':
case 'O': countO++; break;
case ' ': countSpace++; break;
case '.': countDot++; break;
}
}
System.out.printf("%s%d%n%s%d%n%s%d%n%s%d", "A: ", countA, "O: ", countO, "Space: ", countSpace, "Dot: ", countDot);
}
输出：
A: 1
O: 4
Space: 8
Dot: 1
于 2017-05-21T06:26:39.217 回答
1
查找重复项的三种方法
public class WAP_PrintDuplicates {
public static void main(String[] args) {
String input = "iabccdeffghhijkkkl";
findDuplicate1(input);
findDuplicate2(input);
findDuplicate3(input);
}
private static void findDuplicate3(String input) {
HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
for (int i = 0; i < input.length() - 1; i++) {
int ch = input.charAt(i);
if (hm.containsKey(input.charAt(i))) {
int value = hm.get(input.charAt(i));
hm.put(input.charAt(i), value + 1);
} else {
hm.put(input.charAt(i), 1);
}
}
Set<Entry<Character, Integer>> entryObj = hm.entrySet();
for (Entry<Character, Integer> entry : entryObj) {
if (entry.getValue() > 1) {
System.out.println("Duplicate: " + entry.getKey());
}
}
}
private static void findDuplicate2(String input) {
int i = 0;
for (int j = i + 1; j < input.length(); j++, i++) {
if (input.charAt(i) == input.charAt(j)) {
System.out.println("Duplicate is: " + input.charAt(i));
}
}
}
private static void findDuplicate1(String input) {
// TODO Auto-generated method stub
for (int i = 0; i < input.length(); i++) {
for (int j = i + 1; j < input.length(); j++) {
if (input.charAt(i) == input.charAt(j)) {
System.out.println("Duplicate is: " + input.charAt(i));
}
}
}
}
}
于 2017-07-26T11:16:26.817 回答
0
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
public class DuplicateCountChar{
public static void main(String[] args) {
Scanner inputString = new Scanner(System.in);
String token = inputString.nextLine();
char[] ch = token.toCharArray();
Map<Character, Integer> dupCountMap = new HashMap<Character,Integer>();
for (char c : ch) {
if(dupCountMap.containsKey(c)) {
dupCountMap.put(c, dupCountMap.get(c)+1);
}else {
dupCountMap.put(c, 1);
}
}
for (char c : ch) {
System.out.println("Key = "+c+ "Value : "+dupCountMap.get(c));
}
Set<Character> keys = dupCountMap.keySet();
for (Character character : keys) {
System.out.println("Key = "+character+ " Value : " + dupCountMap.get(character));
}
}**
于 2016-05-31T04:24:56.020 回答
0
在java中...使用for循环：
import java.util.Scanner;
/**
*
* @author MD SADDAM HUSSAIN */
public class Learn {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
String input = sc.next();
char process[] = input.toCharArray();
boolean status = false;
int index = 0;
for (int i = 0; i < process.length; i++) {
for (int j = 0; j < process.length; j++) {
if (i == j) {
continue;
} else {
if (process[i] == process[j]) {
status = true;
index = i;
break;
} else {
status = false;
}
}
}
if (status) {
System.out.print("" + process[index]);
}
}
}
}
于 2016-06-18T09:52:20.380 回答
0
public class StringCountwithOutHashMap {
public static void main(String[] args) {
System.out.println("Plz Enter Your String: ");
Scanner sc = new Scanner(System.in);
String s1 = sc.nextLine();
int count = 0;
for (int i = 0; i < s1.length(); i++) {
for (int j = 0; j < s1.length(); j++) {
if (s1.charAt(i) == s1.charAt(j)) {
count++;
}
}
System.out.println(s1.charAt(i) + " --> " + count);
String d = String.valueOf(s1.charAt(i)).trim();
s1 = s1.replaceAll(d, "");
count = 0;
}}}
于 2016-11-04T05:42:58.630 回答
0
public class CountH {
public static void main(String[] args) {
String input = "Hi how are you";
char charCount = 'h';
int count = 0;
input = input.toLowerCase();
for (int i = 0; i < input.length(); i++) {
if (input.charAt(i) == charCount) {
count++;
}
}
System.out.println(count);
}
}
于 2017-03-05T17:41:22.473 回答
0
public class DuplicateValue {
public static void main(String[] args) {
String s = "hezzz";
char []st=s.toCharArray();
int count=0;
Set<Character> ch=new HashSet<>();
for(Character cg:st){
if(ch.add(cg)==false){
int occurrences = Collections.frequency(ch, cg);
count+=occurrences;
if(count>1){
System.out.println(cg + ": This character exist more than one time");
}
else{
System.out.println(cg);
}
}
}
System.out.println(count);
}
}
于 2017-03-25T18:49:34.257 回答
0
Map<Character,Integer> listMap = new HashMap<Character,Integer>();
Scanner in= new Scanner(System.in);
System.out.println("enter the string");
String name=in.nextLine().toString();
Integer value=0;
for(int i=0;i<name.length();i++){
if(i==0){
listMap.put(name.charAt(0), 1);
}
else if(listMap.containsKey(name.charAt(i))){
value=listMap.get(name.charAt(i));
listMap.put(name.charAt(i), value+1);
}else listMap.put(name.charAt(i),1);
}
System.out.println(listMap);
于 2017-06-28T22:29:38.277 回答
0
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String reverse1;
String reverse2;
int count = 0;
while(n > 0)
{
String A = sc.next();
String B = sc.next();
reverse1 = new StringBuffer(A).reverse().toString();
reverse2 = new StringBuffer(B).reverse().toString();
if(!A.equals(reverse1))
{
for(int i = 0; i < A.length(); i++)
{
for(int j = 0; j < A.length(); j++)
{
if(A.charAt(j) == A.charAt(i))
{
count++;
}
}
if(count % 2 != 0)
{
A.replace(A.charAt(i),"");
count = 0;
}
}
System.out.println(A);
}
n--;
}
}
}
于 2017-09-01T17:19:09.410 回答
0
public class list {
public static String name(Character k){
String s="the quick brown fox jumped over the lazy dog.";
int count=0;
String l1="";
String l="";
List<Character> list=new ArrayList<Character>();
for(int i1=0;i1<s.length();i1++){
list.add(s.charAt(i1));
}
list.sort(null);
for (Character character : list) {
l+=character;
}
for (int i1=0;i1<l.length();i1++) {
if((l.charAt(i1)==k)){
count+=1;
l1=l.charAt(i1)+" "+Integer.toString(count);
if(k==' '){
l1="Space"+" "+Integer.toString(count);
}
}else{
count=0;
}
}
return l1;
}
public static void main(String[] args){
String g = name('.');
System.out.println(g);
}
}
于 2018-04-01T03:32:10.113 回答
0
查找字符出现的简单方法 >
void findOccurrences() {
String s = "The quick brown fox jumped over the lazy dog.";
Map<String, Integer> occurrences = new LinkedHashMap<String, Integer>();
for (String ch : s.split("")) {
Integer count = occurrences.get(ch);
occurrences.put(ch, count == null ? 1 : count + 1);
}
System.out.println(occurrences);
}
这将打印输出为：
{T=1, h=2, e=4, =8, q=1, u=2, i=1, c=1, k=1, b=1, r=2, o=4, w= 1, n=1, f=1, x=1, j=1, m=1, p=1, d=2, v=1, t=1, l=1, a=1, z=1, y=1，g=1，.=1}
于 2021-06-07T13:03:24.327 回答
0
String str = "anand";
Map<Character, Integer> map
= new HashMap<Character, Integer>();
// Converting string into a char array
char[] charArray = str.toCharArray();
for (char c : charArray) {
if (map.containsKey(c)) {
// If character is present increment count by 1
map.put(c, map.get(c) + 1);
}
else {
// If character is not present
//putting this character into map with 1 as it's value.
map.put(c, 1);
}
}
for (Map.Entry<Character, Integer> entry :
map.entrySet()) {
System.out.println(entry.getKey()
+ " : "
+ entry.getValue());
}
输出：
a:2 n:2 d:1
于 2022-02-09T12:28:06.120 回答
-1
public static void main(String[] args) {
char[] array = "aabsbdcbdgratsbdbcfdgs".toCharArray();
char[][] countArr = new char[array.length][2];
int lastIndex = 0;
for (char c : array) {
int foundIndex = -1;
for (int i = 0; i < lastIndex; i++) {
if (countArr[i][0] == c) {
foundIndex = i;
break;
}
}
if (foundIndex >= 0) {
int a = countArr[foundIndex][1];
countArr[foundIndex][1] = (char) ++a;
} else {
countArr[lastIndex][0] = c;
countArr[lastIndex][1] = '1';
lastIndex++;
}
}
for (int i = 0; i < lastIndex; i++) {
System.out.println(countArr[i][0] + " " + countArr[i][1]);
}
}
于 2016-03-05T15:44:47.723 回答
-1
//sample Input
/*2
7
saska
toro
winn
toro
vanco
saska
toro
3
edddddd
edddddd
edddddd*/
//sample output
/*4
1*/
import java.util.ArrayList;
import java.util.Scanner;
public class MyTestWhere {
/**
* @param args
*/
public static void main(String[] args) {
int count, line;
Scanner sn = new Scanner(System.in);
count = sn.nextInt();
sn.nextLine();
for (int i = 0; i < count; i++) {
line = sn.nextInt();
sn.nextLine();
// String numArr[] = new String[line];
ArrayList<String> Arr=new ArrayList<String>();
String first = sn.nextLine();
Arr.add(first);String f;
for (int j = 1; j < line; j++) {
f= sn.nextLine();
for(int k=0;k<Arr.size();k++){
if(f.equalsIgnoreCase(Arr.get(k)))break;
else if(k== (Arr.size()-1)){Arr.add(f);}
}
}
System.out.println(Arr.size());
}
}
}
于 2016-05-13T06:33:21.127 回答
-1
import java.util.Scanner;
class Test
{
static String s2="";
int l;
void countDuplicateCharacters(String Str)
{
String S=Str.toLowerCase();
for(int i=0;i<S.length();i++)
{
int k=1;
boolean value= repeatedCheck(S.charAt(i));
if(value==true)
continue;
for(int j=i+1;j<S.length();j++)
{
if(S.charAt(i)==S.charAt(j))
{ k++;
}
}
System.out.println("character '" +S.charAt(i)+"' : "+k);
s2=s2+S.charAt(i);
}
}
boolean repeatedCheck(char ch)
{
l=s2.length();
for (int i=0;i<l;i++)
{
if(s2.charAt(i)==ch)
{
return true;
}
}
return false;
}
}
public class Duplicacy {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("Enter any String");
String s=sc.nextLine();
Test t=new Test();
t.countDuplicateCharacters(s);
}}
于 2018-04-19T05:00:18.353 回答
-1
我用二维数组解决了这个问题。输入：aaaabbbcdefggggh 输出：a4b3cdefg4h
int[][] arr = new int[10][2];
String st = "aaaabbbcdefggggh";
char[] stArr = st.toCharArray();
int i = 0;
int j = 0;
for (int k = 0; k < stArr.length; k++) {
if (k == 0) {
arr[i][j] = stArr[k];
arr[i][j + 1] = 1;
} else {
if (arr[i][j] == stArr[k]) {
arr[i][j + 1] = arr[i][j + 1] + 1;
} else {
arr[++i][j] = stArr[k];
arr[i][j + 1] = 1;
}
}
}
System.out.print(arr.length);
String output = "";
for (int m = 0; m < arr.length; m++) {
if (arr[m][1] > 0) {
String character = Character.toString((char) arr[m][0]);
String cnt = arr[m][1] > 1 ? String.valueOf(arr[m][1]) : "";
output = output + character + cnt;
}
}
System.out.print(output);
于 2020-02-10T13:43:35.507 回答
-2
public static void findDuplicate(String letter)
{
for(int i=0; i<letter.length();i++)
{
for( int j=i+1; j<letter.length();j++)
{
if(letter.charAt(i)==letter.charAt(j))
{
System.out.println(letter.charAt(j));
break;
}
}
}
}
findDuplicate("JAVA");
输出是：A
于 2017-06-10T21:25:53.210 回答