您可以使用模运算子（%）来检查字串长度是否为偶数：

string.length() % 2 == 0

要完成这个过程，你可以使用Arrays.stream()来查找长度为偶数的最长字符串：

String longestWord = Arrays.stream(sentence.split(" ")) // creates the stream with words
        .filter(s -> s.length() % 2 == 0) // filters only the even length strings
        .max(Comparator.comparingInt(String::length)) // finds the string with the max length
        .orElse(" "); // returns " " if none string is found

变化是，您比较〉与最长长度，而不是〉=，并检查长度是否为偶数。
若要适应存在其他符号（如.""）的情况，请使用Regex模式。

public static void main(String[] args) {
    // TODO Auto-generated method stub

    String input = "I am good.";
    String[] input_words = input.split(" ");
    String longestWord = " ";

    for(String word : input_words) {
        Pattern p = Pattern.compile("^[a-zA-Z]+");
        Matcher m = p.matcher(word);
        m.find();
        if(m.group().length() % 2 == 0 && m.group().length() > longestWord.length()) {
            longestWord = m.group();
        }
    }
    System.out.println("result : " + longestWord);
}

这将打印第一个出现的最大偶数字

使用%（模数）运算符是通过检查除以2是否得出余数来评估数字是奇数还是偶数的常用方法。在这里可以这样使用它：

if (array[i].length() >= longestWord.length() && array[i].length() % 2 == 0) {
                longestWord = array[i];
         }

编辑：我感觉很糟糕，因为这听起来像是一项作业，所以我不会解决问题中“先来”的部分。

public class HelloWorld {

     public static void main(String []args) {

        String sentence = "this is a sample input for the problem";
        
        String arr[] = sentence.split("\\s+"); // to split the sentence by space (because words are separated by space) 
        
        int len =  0; 
        String word = "";
        for (int i= 0; i< arr.length; i++) {
            if((arr[i].length())%2 == 0) { // check if the length of the word is even 
                int len1 = arr[i].length();
                if (len1 > len ) { // if new length is greater than the previous word's length then assign the value of new length to len variable
                    len = len1;
                    word = arr[i];  // word = highest length word
                }
            }
        }
        System.out.println(word);
     }
}

python中问题的解决方案是：

def longestevenword(sentence):
  #converting the sentence into lists
  string_list = sentence.split()
  size = 0
  #iteration through each word in the list
  for word in string_list:
    #check to see if the word has even number of characters
    if len(word)%2 == 0:
      #checking if the length of the of the word
      if size <= len(word):
        size = len(word)
        even_word = word
    else:
      continue        
  # printing the longest even word in a given sentence.
  print(even_word)

## function call to test
longestevenword("This is a cat that had a puppyy")

Output: puppyy

我的GitHub https://github.com/jaikishpai/CommonSolutions/blob/main/LongestEvenWord.py上也有

public static void main（String[] args）{ // TODO自动生成的方法存根

String input = "I am good.";
String[] input_words = input.split(" ");
String longestWord = " ";

for(String word : input_words) {
    Pattern p = Pattern.compile("^[a-zA-Z]+");
    Matcher m = p.matcher(word);
    m.find();
    if(m.group().length() % 2 == 0 && m.group().length() > longestWord.length()) {
        longestWord = m.group();
    }
}
System.out.println("result : " + longestWord);

}

public class LongestEvenString {
    public static void main(String[] args) {
        String test = "this is a sample input for the problem";
        System.out.println(longestEvenString(test));
    }

    public static String longestEvenString(String test) {
//        Splits a sentence to array of Strings.
        String[] words = test.split(" ");
//        curlen stores length of current word in string Array
//        maxlen stores length of maximum_length word in string Array
        int curlen = 0;
        int maxlen = 0;
        String output = "";
        for (String word:words) {
            curlen = word.length();
//            loop runs only if length of the current word is more than
//            maximum_length thus excluding future same maximum_length words.
//            and also checks for even_length of that word
            if(curlen > maxlen && curlen%2 == 0){
                maxlen = curlen;
                output = word;
            }
        }
//        returns maximum_even_length word which occurred first
        return output;
    }
}

公共类FindLongestEvenWord {

public static void main(String[] args) {
    String s = "Be not afraid of greatness some are born great some achieve greatness";
    
    List<String> strList = Arrays.asList(s.trim().split(" "));
    Optional<String>  maxLengthEvenWord = strList.stream().filter(s3->s3.length()%2==0).reduce((s1,s2)->s1.length()>s2.length()?s1:s2);
    if(maxLengthEvenWord.isPresent())
        System.out.println(maxLengthEvenWord.get());
    
}

}