将字符串中的字母提升为java中的下一个字母

我有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。不使用布尔值来做到这一点。 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az的小写字母。 任何帮助将不胜感激谢谢。

Answer 1

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’因此给出25 + 1 = 26，模数26 = 0.然后再添加’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); 

会输出

 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即可。 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：

** 免责声明 **：包含您的作业解决方案

 // 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 az 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))); 

会输出

 l hemmo wormd 

** 编辑2 **

如果你有一个基于索引的字母表（如果你需要包含额外的字符等），这里是另一个解决方案。 没有评论也没有优化，但代码有效且应该是自我解释的……仅限FYI：

 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= 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 

Answer 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); 

Answer 3

又一个选项，将== test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 + " "); } } 

Answer 4

 Map s = new HashMap(); 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) 

;

Answer 5

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，你不能去下一个字符，所以我把它做到了开头，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 az, 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()) ; } } 

Answer 6

可能对某人有用：

 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();