我需要在字符串中找到一個(gè)既長又均勻的單詞。舉幾個(gè)例子：這句話.這應(yīng)該返回字符串，因?yàn)樗亲畲箝L度的單詞，長度為 ，而不是因?yàn)闃?gòu)造是奇數(shù)的長度。Time to construct great arttimeeven4construct9另外，在 的示例中。這應(yīng)該返回字符串 ，因?yàn)樗桥紨?shù)，并且偶數(shù)長度為 。它不會(huì)返回單詞，因?yàn)槭紫瘸霈F(xiàn)。Time to write great codeTime4codeTime String[] array = sentence.split(" "); String longestWord = " "; for (int i = 0; i < array.length; i ++) { if (array[i].length() >= longestWord.length()) { longestWord = array[i]; } } System.out.println(longestWord);我的代碼成功地打印了最長的單詞，但是沒有考慮最長的字符串長度是否為偶數(shù)，以及它是否首先出現(xiàn)。我嘗試過在我的for循環(huán)中使用一些模數(shù)字符，但它并沒有跟蹤最偉大的單詞是否甚至移動(dòng)到下一個(gè)最大的單詞。另外，我很難跟蹤這個(gè)詞是否先出現(xiàn)。我試圖解釋甚至：for (int i = 0; i < array.length; i ++) { if (array[i].length() >= longestWord.length()) { longestWord = array[i]; if (longestWord.length() % 2 != 0) { break; } else { longestWord = array[i]; } } }
4 回答
RISEBY
TA貢獻(xiàn)1856條經(jīng)驗(yàn) 獲得超5個(gè)贊
您可以使用模運(yùn)算符 （） 來檢查字符串長度是否為偶數(shù)：%
string.length() % 2 == 0
為此，您可以使用 Arrays.stream（） 來查找長度為偶數(shù)的最長字符串：
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
www說
TA貢獻(xiàn)1775條經(jīng)驗(yàn) 獲得超8個(gè)贊
更改是您比較最長長度的>，而不是>=并檢查長度是否均勻。
要適應(yīng)存在其他符號(hào)（如 '.'）的情況，請使用正則表達(dá)式模式。
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);
}
這將打印出最大的第一個(gè)偶數(shù)字
桃花長相依
TA貢獻(xiàn)1860條經(jīng)驗(yàn) 獲得超8個(gè)贊
在python中這個(gè)問題的解決方案是：
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
它也可以在我的GitHub https://github.com/jaikishpai/CommonSolutions/blob/main/LongestEvenWord.py
jeck貓
TA貢獻(xiàn)1909條經(jīng)驗(yàn) 獲得超7個(gè)贊
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;
}
}
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