我们给出了一个字符串，例如，取“TUOPPPPJHHTT” 我们希望找出哪个字符出现的次数最多，连续在字符串中出现多少次。 在这种情况下，其P发生4次。字符串中最重复的字符
我试图for循环运行如下
char[] array = S.toCharArray();
int count=1;
for(int i =1; i < S.length(); i++) {
if(array[i] == array[i-1]) {
count++;
}
}
但在这种方法中，问题是，它会计算所有字母的重复出现。
每次找到与前一个字符不同的字符时，表示运行（连续重复字母）结束，因此应记下当前运行的长度（即值为count），然后重置计数。最后你可以打印最大值。
char[] array = S.toCharArray()
int count = 1;
int max = 0;
char maxChar = 0;
for(int i=1; i<array.length; i++){ // Start from 1 since we want to compare it with the char in index 0
if(array[i]==array[i-1]){
count++;
} else {
if(count>max){ // Record current run length, is it the maximum?
max=count;
maxChar=array[i-1];
}
count = 1; // Reset the count
}
}
if(count>max){
max=count; // This is to account for the last run
maxChar=array[array.length-1];
}
System.out.println("Longest run: "+max+", for the character "+maxChar); // Print the maximum.
试试这个，
char[] array = input.toCharArray();
Arrays.sort(array);
int max = 0;
int count = 1;
char temp = array[0];
for (char value : array)
{
if (value == temp)
{
count++;
}
else
{
temp = value;
if (count > max)
{
max = count;
}
count = 1;
}
}
if(count > max)
{
max = count;
}
System.out.println("count : "+max);
这里有一个更通用的解决方案，为所有文字作品;字母数字或特殊字符无关紧要。
private String findMaxChar(String str) {
char[] array = str.toCharArray();
int maxCount = 1;
char maxChar = array[0];
for(int i = 0, j = 0; i < str.length() - 1; i = j){
int count = 1;
while (++j < str.length() && array[i] == array[j]) {
count++;
}
if (count > maxCount) {
maxCount = count;
maxChar = array[i];
}
}
return (maxChar + " = " + maxCount);
}
System.out.println(findMaxChar("T"));
System.out.println(findMaxChar("TDD"));
System.out.println(findMaxChar("WWW"));
System.out.println(findMaxChar("NOREPEATS"));
System.out.println(findMaxChar("122333444455555"));
System.out.println(findMaxChar("abc33++$$$_###*ABCC"));
输出：
T = 1
D = 2
W = 3
N = 1 // First Character (if no repeats)
5 = 5
$ = 3
Set收集它们为：
private static String findMaxChar(String str) {
char[] array = str.toCharArray();
Set<Character> maxChars = new LinkedHashSet<Character>();
int maxCount = 1;
maxChars.add(array[0]);
for(int i = 0, j = 0; i < str.length() - 1; i = j){
int count = 1;
while (++j < str.length() && array[i] == array[j]) {
count++;
}
if (count > maxCount) {
maxCount = count;
maxChars.clear();
maxChars.add(array[i]);
} else if (count == maxCount) {
maxChars.add(array[i]);
}
}
return (maxChars + " = " + maxCount);
}
输出：
[T] = 1
[D] = 2
[W] = 3
[N, O, R, E, P, A, T] = 1
[5] = 5
[$, #] = 3 // All Characters (in case of a tie)
请试试这个代码我做到了，我.........
public static String getMaxRepeatedChar(String txt) {
if ((txt != null)) {
HashMap<Character, Integer> hash = new HashMap<Character, Integer>();
char maxCh = 1;
int maxCnt = 0;
for (char ch : txt.toCharArray()) {
if (hash.containsKey(ch)) {
int i = hash.get(ch);
hash.put(ch, i + 1);
if (maxCnt < (i + 1)) {
maxCh = ch;
maxCnt = 1 + i;
}
} else {
hash.put(ch, 1);
if (maxCnt < 1) {
maxCh = ch;
maxCnt = 1;
}
}
}
return "Most Repeted character : " + maxCh + " and Count : "
+ maxCnt;
}
return null;
}
它在单个循环中做出了答案........ –
您可以使用三元运算创建的简化版本@ justhalf的解决方案。另外，如果使用'charAt（）'方法，则不需要先将字符串转换为字符数组。
int count = 1;
int max = 1;
char maxChar = S.charAt(1);
for(int i = 1; i < S.length(); i++) {
count = (S.charAt(i) == S.charAt(i - 1)) ? (count + 1) : 1;
if (count > max) {
max = count;
maxChar = S.charAt(i);
}
}
System.out.println("Longest run: "+max+", for the character "+maxChar);
请注意，此代码假定S不是空的。
private static void findMaxChar(string S)
{
char[] array = S.ToLower().ToCharArray();
int count = 1;
int max = 0;
char maxChar = '0';
for (int i = 0; i < S.Length-1; i++)
{ // Note that it should be S.length instead of S.length()
string stringleft=S.Replace(array[i].ToString(),"");
int countleft = S.Length - stringleft.Length;
count = countleft;
if (count > max)
{ // Record current run length, is it the maximum?
max = count;
maxChar = array[i];
}
}
// This is to account for the last run
Console.WriteLine("Longest run: "+max+", for the character "+maxChar);
}
这只是代码的增强，可以在csharp中使用 –
str = "AHSDFHLFHHSKSHDKDHDHHSJDKDKSDHHSDKKSDHHSDKLLDFLDFLDHAFLLLFFASHHHHHHYYYYYYYYYAAAAAAASDFJK"
max = 1
fin_max =1
for i in range(0,len(str)-1):
if(str[i]==str[i+1]):
max = max + 1
if fin_max < max:
fin_max = max
else:
max = 1
print fin_max
import java.util.*;
public class HighestOccurence {
public static void getHighestDupsOccurrancesInString(char[] arr) {
int count = -1;
int max = 0;
Character result = ' ';
// Key is the alphabet and value is count
HashMap<Character, Integer> hmap = new HashMap<Character, Integer>();
for (int i = 0; i < arr.length; i++) {
if (hmap.containsKey(arr[i])) {
hmap.put(arr[i], hmap.get(arr[i]) + 1);
} else {
hmap.put(arr[i], 1);
}
}
for (Map.Entry<Character, Integer> itr : hmap.entrySet()) {
// System.out.println(itr.getKey() + " " + itr.getValue());
if (Integer.parseInt(itr.getValue().toString()) > max) {
max = Integer.parseInt(itr.getValue().toString());
if (hmap.containsValue(max)) {
result = itr.getKey();
}
}
}
System.out.print("Maximum Occured Character is '" + result + "'");
System.out.print(" with count :" + max);
}
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter the String");
String s = scan.nextLine();
if (s.trim().isEmpty()) {
System.out.println("String is empty");
} else {
char[] arr = s.toCharArray();
getHighestDupsOccurrancesInString(arr);
}
}
}
请不要添加代码。尝试添加一个解释。 –
public static int findMaximumRepeatedChar(String str){
int count = 0, maxCount = 0, charAt = -1;
for(int i=0; i < str.length() - 1; i++){
if(str.charAt(i) != str.charAt(i+1)){
if(count > maxCount){
maxCount = count;
charAt = i - count + 1;
}
count = 0;
}else{
count++;
}
}
if(charAt == -1) return -1;
else return charAt;
}
对于简单的字符串操作，这个程序是可以做到：
package abc;
import java.io.*;
public class highocc
{
public static void main(String args[])throws IOException
{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter any word : ");
String str=in.readLine();
str=str.toLowerCase();
int g=0,count;
int ar[]=new int[26];
char ch[]={'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'};
for(int i=0;i<ch.length;i++)
{
count=0;
for(int j=0;j<str.length();j++)
{
char ch1=str.charAt(j);
if(ch[i]==ch1)
count++;
}
ar[i]=(int) count;
}
int max=ar[0];
for(int j=1;j<26;j++)
{
if(max<ar[j])
{
max=ar[j];
g=j;
}
else
{
max=ar[0];
g=0;
}
}
System.out.println("Maximum Occurence is "+max+" of character "+ch[g]);
}
}
import java.util.*;
class findmax
{
public static void main(String args[])
{
String s;
int max=0;
Scanner sc=new Scanner(System.in);
System.out.println("Enter String:");
s=sc.next();
String s1=" ";
for(int i=0;i<s.length();i++)
{
int count=0;
for(int j=i+1;j<s.length();j++){
if(s.charAt(i)==s.charAt(j))
count++;
}
if(count>max){
s1=Character.toString(s.charAt(i));
max=count;
}
else if(count==max)
s1=s1+" "+Character.toString(s.charAt(i));
}
System.out.println(s1);
}
}
public static char findMaximum(String input){
int maxCount=0;
int count=1;
char result='\0';
char[] temp=input.toCharArray();
if(temp.length==1){
return temp[0];
}
for(int i=1;i<temp.length;i++){
if(temp[i]==temp[i-1]){
count++;
}
else{
if(count>maxCount){
maxCount=count;
result=temp[i-1];
}
count=1;
}
}
return result;
}
虽然这可能回答OP的问题，但最好的答案是也包含解释的答案。你能否记录你的代码中发生了什么？ – Tgsmith61591
你可以试试这个代码：
import java.util.Scanner;
public class HighestOccuringConsecutiveChar
{
public static char highestOccuringConsecutiveCharacter(String a)
{
int c= 0, cNext = 0;
char ans = '\0';
for(int i = 0; i < a.length()-1; i++)
{
if(a.charAt(i) == a.charAt(i+1))
cNext++;
if(cNext > c)
{
ans = a.charAt(i);
c = cNext;
}
if(a.charAt(i) != a.charAt(i))
cNext = 0;
if(c == 0 && cNext == 0)
ans = a.charAt(i);
}
return ans;
}
public static void main(String[] args)
{
Scanner s = new Scanner(System.in);
String str = new String();
str = s.nextLine();
System.out.println(highestOccuringConsecutiveCharacter(str));
}
}
package naresh.java;
import java.util.HashMap;
import java.util.Map;
public class StringPrg {
public static void main(String args[]){
String str= "Ramrrnarmmmesh";
//using hashmap to store unique character with integer count
Map<Character,Integer> map1 = new HashMap<Character,Integer>();
for(int k=0; k < str.length(); k++)
{
char currentChar = str.charAt(k);
//to check that currentChar is not in map, if not will add 1 count for firsttime
if(map1.get(currentChar) == null){
map1.put(currentChar, 1);
}
/*If it is repeating then simply we will increase the count of that key(character) by 1*/
else {
map1.put(currentChar, map1.get(currentChar) + 1);
}
}
//Now To find the highest character repeated
int max=0;
char maxCharacter = 'a';//setting to a by default
for (Map.Entry<Character, Integer> entry : map1.entrySet())
{
System.out.println("Key=" + entry.getKey() + ":Value" + entry.getValue());
if(max<entry.getValue()){
max=entry.getValue();
maxCharacter=entry.getKey();
}
}
System.out.println("Max Character=" + maxCharacter + "Max Count" + max);
}
}
虽然这个答案确实有帮助，但通过解释你做了什么以及为什么，你可以大大提高答案的质量。 – Zabuza
感谢您的宝贵意见，我在代码中添加了评论，我希望它会有所帮助，还有任何改进的余地，欢迎您的光临！ –
'“什么想法？”' - 其一，缩进和格式化你的代码更好，所以，它的可读性。对于另一种情况，请考虑使用26个项目的数组，每个字母一个，并在关联的插槽中设置最大连续计数。另一个选择是使用'HashMap'。 –
想法：首先找出所有重复的人物序列，然后寻找你找到的最大的人物。 –
你会把大小写成同一个字母吗？ – KeepCalmAndCarryOn