Java Regular Expression: removes spaces (including width and width) on both sides of the string. spaces inside the string cannot be removed]

Remove spaces (including width and width) on both sides of the string. spaces inside the string cannot be removed.

 

I have tried to use only Java instead of Java regular expressions. lang. although the APIS provided in string can be implemented, you need to write a bunch of code. Now you will take a note of the code that does not use Java regular expressions or Java regular expressions.

 

A does not use Java regular expressions. It is a little troublesome. ^_^

Public class spacechecker {</P> <p> Public static void main (string [] ARGs) {<br/> string STR = "a B C "; </P> <p> system. out. println (delspace (STR); <br/>}</P> <p> Public static string delspace (string Str) {</P> <p> If (STR = NULL) {<br/> return NULL; <br/>}</P> <p> // Delete the halfwidth space first <br/> STR = Str. trim (); </P> <p> while (Str. startswith ("") {</P> <p> // unfortunately, the string does not provide replacelast (), otherwise, it will be simpler. <br/> // after this loop is completed, you can only Spaces are deleted, but backend spaces cannot be deleted. <br/> // after this loop is completed, the string is flipped and then spaces are removed. <br/> STR = Str. replacefirst ("", ""); </P> <p> // trim () is required. Otherwise, if the front-end space is full-width and half-width, it cannot be fixed. <br/> STR = Str. trim (); <br/>}</P> <p> // flip the string <br/> STR = reverse (STR ); </P> <p> // enter a space again <br/> while (Str. startswith ("") {</P> <p> STR = Str. replacefirst ("", ""); </P> <p> STR = Str. trim (); <br/>}</P> <p> // Finally, flip the string back. <br/> return STR = reverse (STR ). trim (); <br/>}</P> <P> // custom string flip method. <Br/> // many third-party packages are implemented, but Java APIs are not implemented. Implement ^ _ ^ here. <br/> Public static string reverse (string Str) {<br/> char [] charsold = Str. tochararray (); </P> <p> char [] charsnew = new char [charsold. length]; </P> <p> int Index = charsold. length-1; </P> <p> for (INT I = 0; I <charsold. length; I ++) {</P> <p> charsnew [I] = charsold [index-I]; <br/>}</P> <p> return string. valueof (charsnew); <br/>}< br/>}

B. Use a Java regular expression.

Public class spacechecker {</P> <p> Public static void main (string [] ARGs) throws exception {<br/> spacecheck SC = new spacecheck (); </P> <p> string STR = "a B C"; </P> <p> system. out. println (SC. delspace (STR); <br/>}</P> <p> Public String delspace (string Str) throws exception {</P> <p> If (STR = NULL) {<br/> return NULL; <br/>}</P> <p> string regstartspace = "^ [] *"; <br/> string regendspace = "[] * $ "; </P> <p> // two replaceall in a row <br/> // The first one is to remove the leading space, the second one is to remove the backend space <br/> // originally intended to be in one breath, but it cannot be implemented ^ _ ^ <br/> string strdelspace = Str. replaceall (regstartspace ,""). replaceall (regendspace, ""); </P> <p> return strdelspace; <br/>}< br/>}

 

Take a note and remember to study Java regular expressions.