英文:
Write the code to process the list of words and trim any spaces out of the words
问题
这段代码应该检查字符串数组中的每个字符串，看看单词之间是否有空格。以下是一个示例：{“every”，“near ing”，“checking”，“food”，“stand”，“value”}。它应该被更改为：{“every”，“nearing”，“checking”，“food”，“stand”，“value”}。以下是迄今为止的我的代码：
package space;import java.util.regex.Matcher;import java.util.regex.Pattern;public class spacefinder {public static void main (String[] args) {String[] arr = {"every", "near ing", "checking", "food", "stand", "value"};Pattern pattern = Pattern.compile("\\s");for (int i = 0; i < arr.length; i++) {Matcher matcher = pattern.matcher(arr[i]);arr[i] = matcher.replaceAll("");}}}
这段代码会产生一个错误，而且我看到的每个教程都没有使用字符串数组，比如我这种情况下的 String[] Arr，他们只使用了一个常规的 String = 语句。
英文:
This code is supposed to check through an array of strings and see if there is a space between a word. Here is an example: {“every”, “near ing”, “ checking”, “food “, “stand”, “value “}. It should be changed to hold: {“every”, “nearing”, “checking”, “food”, “stand”, “value”}. Here is my code so far:
package space;import java.util.regex.Matcher;import java.util.regex.Pattern;public class spacefinder {public static void main (String[] args) {String[] arr = {"every", "near ing", "checking", "food", "stand", "value"};Pattern pattern = Pattern.compile("\\arr");Matcher matcher = pattern.matcher(arr);boolean found = matcher.find();}}
This code results in an error, and every tutorial I see does not use a string array, such as the String[] Arr in my case, they only use a regular String = statement.
答案1
得分: 1
你可以尝试类似这样的代码：
for(int i = 0; i < arr.length; i++) {arr[i] = arr[i].replace(" ", "");}
英文:
You can try something like this:
for(int i = 0; i<arr.length; i++) {arr[i] = arr[i].replace(" ", "");}
答案2
得分: -1
也许这会起作用
public static String[] removeSpace(String[] ans) {for(String s:ans){while(s.contains(" ")){s = s.split(" ")[0] + s.split(" ")[1];}}return ans;}
英文:
maby this wil work
public static String[] removeSpace(String[] ans) {for(String s:ans){while(s.contains(" ")){s = s.split(" ")[0] + s.split(" ")[1];}}return ans;}
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