我遇到问题，弄清楚如何让我的代码增加用户输入给出的字符串，这样当用户选择替换像z这样的字母时，它会转到a，b到c等。我必须这样做而不使用布尔值。我应该通过使用算术来从用户输入获得从z到a的促销。加上必须只是a-z的小写字母。任何帮助将不胜感激。
答案 0 :(得分：18)
这段代码
String foo = "abcdefz";
String bar = "";
for (char c : foo.toCharArray()) {
bar += Character.toString((char) (((c - 'a' + 1) % 26) + 'a'));
}
System.out.println(bar);
将输出
bcdefga
它做的是取字符，减去'a'的字符代码，从而得到0到25之间的值。然后我们递增1.得到答案并执行模数26，所以如果我们有'z' ，我们减去'a'，从而给出25 + 1 = 26，模数26 = 0.然后我们再添加'a'并vo！
** 编辑 **
你甚至可以进一步推动这个概念并添加一个变量“移位”值：
int shiftValue = 12;
String foo = "abcdefz";
String bar = "";
for (char c : foo.toCharArray()) {
bar += Character.toString((char) (((c - 'a' + shiftValue) % 26) + 'a'));
}
System.out.println(bar);
将输出
mnopqrl
shiftValue的值可以是任何正整数（即，移位-2与移位24相同）。试试吧。
** 更新 **
好吧，只需用等式替换你的alpha + 1。并不是说我想给你提供一切，但如果你必须坚持，这就是你需要做的事情：
** 免责声明 **：包含您的家庭作业解决方案
// define some constants
char FIRST_LETTER = 'a'; // the first letter in the alphabet
int ALPHABET_SIZE = 26; // the number of letters in the alphabet
int SHIFT_VALUE = 1; // number of letters to shift
Scanner kb = new Scanner(System.in);
String second = "hello world"; // target string
String alphabet = kb.next();
// TODO: need to check if alphabet has at least one char and if it's in the range of a-z
char alpha = alphabet.charAt(0); // just keep the first char in the input
System.out.println(second.replace(alpha, (char) (((alpha - FIRST_LETTER + SHIFT_VALUE) % ALPHABET_SIZE ) + FIRST_LETTER)));
将输出
l
hemmo wormd
** 编辑2 **
如果你有一个基于索引的字母（如果你需要包含额外的字符等），这是另一个解决方案。没有评论也没有优化，但代码有效且应该是自我解释的......仅供参考：
int shiftValue = 1;
char[] alphabet = new char[] {
'a','b','c','d','e','f','g','h','i',
'j','k','l','m','n','o','p','q','r',
's','t','u','v','w','x','y','z','!',' '
};
boolean[] replace = new boolean[alphabet.length];
Scanner kb = new Scanner(System.in);
String text = "hello world !";
System.out.print("$ ");
String input = kb.nextLine().toLowerCase();
Arrays.fill(replace, false);
for (char c : input.toCharArray()) {
int index = -1;
for (int i=0; i<alphabet.length; i++) {
if (alphabet[i] == c) {
index = i;
break;
}
}
if (index >= 0) {
replace[index] = true;
}
}
for (int i=alphabet.length - 1; i>0; i--) {
if (replace[i]) {
text = text.replace(alphabet[i], alphabet[(i+shiftValue) % alphabet.length]);
}
}
System.out.println(text);
当然，此代码将替换text字符串中从 stdin 读取的每个char。输出的一个例子是：
$ ! e wo
hfllpaxprlda
答案 1 :(得分：2)
String s = "q";
char c = s.charAt(0);
c++;
//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);
答案 2 :(得分：2)
又一个选项，将==测试隐藏在交换机中：
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 + " ");
}
}
答案 3 :(得分：1)
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)
答案 4 :(得分：1)
我有以下答案。它处理字符串之间的空白区域。当你有一个'z'，这是字母表的结尾时，你不能去下一个字符，所以我把它带到了开头，这是一个。
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()) ;
}
}
答案 5 :(得分：0)
也许对某人有用：
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) || ((int)c >= 97) && ((int)c <= 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();
答案 6 :(得分：0)
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'
public static String stringRoll(String str, int[] arr) {
char[] charArr = str.toCharArray();
int valueOfZ = 'z';
for(int i = 0; i < arr.length; i++){
for(int j = 0; j < arr[i]; j++){
int value = charArr[j] + 1;
if(value > valueOfZ) {
charArr[j] = 'a';
}else{
charArr[j] = (char) value;
}
}
}
return new String(charArr);
}
答案 7 :(得分：0)
公共类 ReplaceWithNextChar {
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++)
{
val=ch[i]+1;
char c=(char)val;
s1=s1+c;
}
System.out.println(s1);
}
}