blocks|key|1036709|text|String+s+=+"q";
char+c+=+s.charAt(0);
c%2B%2B;
//here+you+can+handle+cases+like+z->a
if+(c=='z')+c+=+'a';

char[]+chars+=+new+char[1];
chars[0]+=+c;
String+s1+=+new+String(chars);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1036710|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>String s = "q";
char c = s.charAt(0);
c++;
//here you can handle cases like z-&gt;a
if (c=='z') c = 'a';

char[] chars = new char[1];
chars[0] = c;
String s1 = new String(chars);
</code></pre>

blocks|key|1036756|text|还有另一种选择，将==测试隐藏在开关中：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1036757|public+static+void+main(String[]+args)
{

++++char[]+chars+=+{'a','k','f','z'};
++++for+(char+letter+:+chars)+{
++++++++char+next+=+letter;
++++++++switch+(next)+{
++++++++case+122:+//+or+'z',+either+way
++++++++++++next+='a';
++++++++++++break;
++++++++default:
++++++++++++next+%2B=+1;
++++++++}
++++++++System.out.println(letter+%2B+"+is+followed+by+"+%2B+next+%2B+"+");
++++}

}|code-block|syntax|javascript|1036758|entityMap^0|9|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

yet another option, hiding the <code>==</code> test in a switch:

<pre><code>public static void main(String[] args)
{

 char[] chars = {'a','k','f','z'};
 for (char letter : chars) {
 char next = letter;
 switch (next) {
 case 122: // or 'z', either way
 next ='a';
 break;
 default:
 next += 1;
 }
 System.out.println(letter + " is followed by " + next + " ");
 }

}
</code></pre>

blocks|key|1036717|text|Map<Character,+Character>+s+=+new+HashMap<Character,+Character>();
s.put('a',+'b');
...
s.put('z',+'a');


String+s+=+"q";
char+c+=+s.charAt(0);

c+=+s.get(c);

char[]+chars+=+new+char[1];
chars[0]+=+c;
String+s1+=+new+String(chars)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1036718|；|unstyled|1036719|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>Map&lt;Character, Character&gt; s = new HashMap&lt;Character, Character&gt;();
s.put('a', 'b');
...
s.put('z', 'a');


String s = "q";
char c = s.charAt(0);

c = s.get(c);

char[] chars = new char[1];
chars[0] = c;
String s1 = new String(chars)
</code></pre>

;

blocks|key|1036806|text|我有以下答案。它负责字符串之间的空格。还有，当你有一个'z'，这是字母表的结尾，你不能转到下一个字符，所以我让它转到开头，也就是a。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1036807|import+java.util.Scanner;

public+class+NextChar+{

++++public+static+char+nextCharacterInAlphabet(char+character)
++++{
++++++++char+nextChar;+++++++++++
++++++++int+ascii+=+(int)+character;

++++++++if+(ascii+==32)++//ascii+code+for+a+space+is+32
++++++++++++/*space+stays+space+that+separates+the+different+
++++++++++++strings+seperated+by+the+spaces+*/
++++++++++++nextChar+=+(char)+ascii;+++
++++++++else+if+(ascii+==+122)
++++++++++++/*if+the+character+is+a+z,+then+there+is+no+
++++++++++++next+character+so+i+let+i+go+back+to+the+first+character+in+alphabet*/
++++++++nextChar+=+(char)+(ascii+-25);+//alphabet+has+26+chars,+z=26+-25+is+first+char
++++++++++++++else+//if+the+char+is+not+a+space+or+z+then+it+goes+to+the+next+char
++++++++++++++++nextChar+=+(char)+(ascii+%2B1);
++++++++return+nextChar;+
++++}

++++public+static+void+main(String[]+args)+{

++++++++Scanner+input+=+new+Scanner(+System.in+);
++++++++StringBuffer+sb+=+new+StringBuffer();+//to+store+the+characters

++++++++for+(char+c+:+input.nextLine().toCharArray()+)+//loops+through+the+char+array
++++++++{
++++++++++++sb.append(nextCharacterInAlphabet(c));+//appends+the+chars+to+the+stringbuffer+passed+to+it+by+the+method+
++++++++}
++++++++System.out.println(sb.toString())+;+
++++}
}|code-block|syntax|javascript|1036808|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I have the following answer. It takes care of the white spaces in between strings of characters. also when you have a 'z', which is the end of the alphabet, you can't go to the next character, so i made it go to the beginning, which is a.

<pre><code>import java.util.Scanner;

public class NextChar {

 public static char nextCharacterInAlphabet(char character)
 {
 char nextChar; 
 int ascii = (int) character;

 if (ascii ==32) //ascii code for a space is 32
 /*space stays space that separates the different 
 strings seperated by the spaces */
 nextChar = (char) ascii; 
 else if (ascii == 122)
 /*if the character is a z, then there is no 
 next character so i let i go back to the first character in alphabet*/
 nextChar = (char) (ascii -25); //alphabet has 26 chars, z=26 -25 is first char
 else //if the char is not a space or z then it goes to the next char
 nextChar = (char) (ascii +1);
 return nextChar; 
 }

 public static void main(String[] args) {

 Scanner input = new Scanner( System.in );
 StringBuffer sb = new StringBuffer(); //to store the characters

 for (char c : input.nextLine().toCharArray() ) //loops through the char array
 {
 sb.append(nextCharacterInAlphabet(c)); //appends the chars to the stringbuffer passed to it by the method 
 }
 System.out.println(sb.toString()) ; 
 }
}
</code></pre>

blocks|key|821734|text|也许它会对某些人有用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|821735|import+java.util.Scanner;
public+static+void+main(String[]+args)+{
++++++++Scanner+input+=+new+Scanner(System.in);
++++++++String+ss+=+new+String();
++++++++System.out.print("Enter+String:");

++++for(char+c+:+input.nextLine().toCharArray()+){
++++++++//+Checking+given+string+must+cantain+only+[a-zA-Z\s]+values+(by++ASCII+Table);
++++++++//+(condition)+char+value+must+be+alphabeticl+value+by+ASCII+Table
++++++++if+(+(+(int)c+>=+65+)+&&+((int)c+<=+90)+%7C%7C+((int)c+>=+97)+&&+((int)c+<=+122)+)+{
++++++++++++switch+((int)c+)+{
++++++++++++++++case+90:
++++++++++++++++++++ss+%2B=+"A";
++++++++++++++++++++break;
++++++++++++++++case+122:
++++++++++++++++++++ss+%2B=+"a";
++++++++++++++++++++break;
++++++++++++++++default:
++++++++++++++++++++ss+%2B=+(char)(c%2B1);
++++++++++++}//+switch()+increment+every+char+by+one

++++++++}else+if(+(int)c+==+32+){
++++++++++++ss+%2B=+"+";
++++++++}else{
++++++++++++ss+%2B=(char)(c);
++++++++}//+if(char+value+must+alphabetical+ASCII+condition);

++++}+//+--For+Loop+

++++System.out.println("\nnew+String:+"%2Bss);

}+//+main();|code-block|syntax|javascript|821736|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

May be it will useful for someone:

<pre><code>import java.util.Scanner;
public static void main(String[] args) {
 Scanner input = new Scanner(System.in);
 String ss = new String();
 System.out.print("Enter String:");

 for(char c : input.nextLine().toCharArray() ){
 // Checking given string must cantain only [a-zA-Z\s] values (by ASCII Table);
 // (condition) char value must be alphabeticl value by ASCII Table
 if ( ( (int)c &gt;= 65 ) &amp;&amp; ((int)c &lt;= 90) || ((int)c &gt;= 97) &amp;&amp; ((int)c &lt;= 122) ) {
 switch ((int)c ) {
 case 90:
 ss += "A";
 break;
 case 122:
 ss += "a";
 break;
 default:
 ss += (char)(c+1);
 }// switch() increment every char by one

 }else if( (int)c == 32 ){
 ss += " ";
 }else{
 ss +=(char)(c);
 }// if(char value must alphabetical ASCII condition);

 } // --For Loop 

 System.out.println("\nnew String: "+ss);

} // main();
</code></pre>

blocks|key|1036874|text|根据数组长度滚动每个字符，当字符值大于'z‘时，将值更改为'a’|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1036875|public+static+String+stringRoll(String+str,+int[]+arr)+{
+++++++char[]+charArr+=+str.toCharArray();
+++++++int+valueOfZ+=+'z';
++++++for(int+i+=+0;+i+<+arr.length;+i%2B%2B){
++++++++++for(int+j+=+0;+j+<+arr[i];+j%2B%2B){
++++++++++++++++int+value+=+charArr[j]+%2B+1;
+++++++++++++++if(value+>+valueOfZ)+{
++++++++++++++++++++charArr[j]+=+'a';
++++++++++++++++}else{
++++++++++++++++++charArr[j]+=+(char)+value;
+++++++++++}
+++++++}
+++}
+++return+new+String(charArr);
}|code-block|syntax|javascript|1036876|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Roll every character based on array length and when character value is greater then 'z' change value to 'a'

<pre><code>public static String stringRoll(String str, int[] arr) {
 char[] charArr = str.toCharArray();
 int valueOfZ = 'z';
 for(int i = 0; i &lt; arr.length; i++){
 for(int j = 0; j &lt; arr[i]; j++){
 int value = charArr[j] + 1;
 if(value &gt; valueOfZ) {
 charArr[j] = 'a';
 }else{
 charArr[j] = (char) value;
 }
 }
 }
 return new String(charArr);
}
</code></pre>

blocks|key|821774|text|公共类ReplaceWithNextChar+{|type|unstyled|depth|inlineStyleRanges|entityRanges|data|821775|public+static+void+main(String[]+args)
{
++Scanner+sc1=new+Scanner(System.in);
++System.out.println("Enter+the+String");
++String+s=sc1.nextLine();
++char[]+ch=s.toCharArray();
++int+val=0;
++String+s1="";
++for(int+i=0;i<ch.length;i%2B%2B)
++{
++++++val=ch[i]%2B1;
++++++char+c=(char)val;
++++++s1=s1%2Bc;
++}
++System.out.println(s1);
}|code-block|syntax|javascript|821776|}|821777|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

public class ReplaceWithNextChar
{
<pre><code>public static void main(String[] args)
{
 Scanner sc1=new Scanner(System.in);
 System.out.println(&quot;Enter the String&quot;);
 String s=sc1.nextLine();
 char[] ch=s.toCharArray();
 int val=0;
 String s1=&quot;&quot;;
 for(int i=0;i&lt;ch.length;i++)
 {
 val=ch[i]+1;
 char c=(char)val;
 s1=s1+c;
 }
 System.out.println(s1);
}
</code></pre>
}

I'm having issues figuring out how to get my code to increment the string that is given by user input so that when a user chooses to replace a letter like z it would go to a, b to c etc. The catch is I have to do this without using boolean. I am supposed to get this by using arithmetics to get the promotion from z to a from the users input. Plus must be only lower case letters from a-z. Any help would be appreciated thanks.

Promoting letters in a string to the next letter in java

我在弄清楚如何让我的代码递增用户输入的字符串时遇到了问题，这样当用户选择替换像z这样的字母时，它将转到a，b到c等等。问题是我必须在不使用布尔值的情况下这样做。我应该通过使用算术从用户输入中获得从z到a的提升来获得此结果。加号只能是a-z的小写字母。任何帮助都将不胜感激。

问将字符串中的字母提升为java中的下一个字母
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回答 7

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问将字符串中的字母提升为java中的下一个字母EN

回答 7

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