Java开发网 - 如何生成x位Ａ

Topic: 如何生成x位Ａ－Ｚ的字串呢？

1.如何生成x位Ａ－Ｚ的字串呢？

Posted by: binge
Posted on: 2006-05-17 12:29

//现有需求如下，生成x位的A-Z的字串，
//x位正整数
//如果为2位，则int x=2;
//生成的字串应该为AA AB AC AE....ZZ
//如果为5位，则int x=5;
//生成的字串应该为AAAAA AAAAB AAAAC AAAAD AAAAE...ZZZZZ
//如果为7位，则int x=7;
//生成的字串应该为AAAAAAA AAAAAAB AAAAAAC AAAAAAD...ZZZZZZZ
//不知是否明白我的意思？

//如果只生成一位可以解决，在两位以上的话就不好解决了，
//要生成A-Z(A到Z之间的所有字符)可以用
for (int i='A';i<='Z';i++){
char atoz=(char)i;
System.out.println(atoz);
}
//谢谢各位

2.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]	Copy to clipboard
Posted by: why Posted on: 2006-05-18 18:15 Are you sure you want to print all those 字串?? x num of 生成的字串 1 26 2 676 3 17576 4 456976 5 11881376 6 308915776 7 8031810176

3.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]	Copy to clipboard
Posted by: xicheng1 Posted on: 2006-05-19 17:12 用无限循环试试 X等于多少就for 几次

4.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]

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Posted by: Jcat
Posted on: 2006-05-21 00:11

Solution1
以下方法的功能：
带入返回
"aaa" "aab"
"aaz" "aba"
"zzz" "aaaa"

public String AddOne(String s) {
        StringBuffer stringBuffer = new StringBuffer(s);
        String re;
        boolean carry = true;
        for (int i = 0; i < stringBuffer.length(); i++) {
            int indexOfTemp = stringBuffer.length() - 1 - i; // begin at the last char
            char temp = stringBuffer.charAt(indexOfTemp);
            if (temp == 'z') {
                stringBuffer.setCharAt(indexOfTemp, 'a');
            } else {
                temp++;
                stringBuffer.setCharAt(indexOfTemp, temp);
                carry = false;
                break;
            }
        }
        if (carry) {
            stringBuffer.insert(0, 'a');
        }
        re = stringBuffer.toString();
        return re;
    }

相信有了这个方法，继续实现你的想法就不难了，剩下的自己搞定吧。

5.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]

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Posted by: Jcat
Posted on: 2006-05-21 00:26

Solution2
比较另类的办法，利用36进制数（Java的Integer类中的最高进制数）的toString方法。
x=2的实现

 public void soCoolAA(int loopNumber) throws Exception{
        if(loopNumber>676){
        throw new Exception("out of range, please make the loopNumber<=676");
        }
        String begin="AA";
        int system36 = Integer.parseInt(begin, 36);
        for (int i = 0; i < loopNumber; i++, system36++) {
            if (i != 0 && i % 26 == 0) {
                system36 += 10;        // skip numbers, since we just need letters
                System.out.print("\n");
            }
            String s = Integer.toString(system36, 36);
            s = s.toUpperCase();
            System.out.print(s + "\t");
        }
    }

调用一下 soCool(676) 看看结果吧。

x＝3的实现

public static void soCoolAAA(int loopNumber) throws Exception {
        if(loopNumber>17576){
            throw new Exception("out of range, please make the loopNumber<=17576");
        }
        String begin="AAA";
        int system36 = Integer.parseInt(begin, 36);
        for (int i = 0; i < loopNumber; i++, system36++) {
            if (i != 0 && i % 26 == 0) {
                system36 += 10;
                System.out.print("\n");
            }
            if (i != 0 && i % (676) == 0) {
                system36 += 36 * 10;
                System.out.println("=====");
            }
            String s = Integer.toString(system36, 36);
            s = s.toUpperCase();
            System.out.print(s + " ");
        }
    }

依此类推吧，不过这个办法效率低下，仅供参考。

6.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]

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Posted by: oujianqiang
Posted on: 2006-05-22 17:53

基础编程题而已，看一下这个程序吧


import java.io.*;
public class randchar
{
  public static void main(String argsp[])throws IOException
  {
    String s;
    int x;
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    System.out.println("insert a number:");
    s=br.readLine();
    x=Integer.parseInt(s);
      for(char j='A';j<='Z';j++)
      { for(char c='A';c<='Z';c++)
       { for(int i=1;i<x;i++)
          System.out.print(j);
         System.out.print(c+"\t");
        }
        System.out.println();
      }
  }
}

{ use [ code ] and check "Disable Smileys" -- why }

7.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]	Copy to clipboard
Posted by: binge Posted on: 2006-05-23 19:31 建议在贴切代码时加入 [ code ] public class A2Z(){ System.out.println("test code reference"); } [/ c"+"o"+"d"+"e ]

8.Re:如何生成x位Ａ－Ｚ的字串呢？ [Re: binge]

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Posted by: chance2000
Posted on: 2006-05-24 18:16

//替换字符串中的index位为newChar
public static String replaceIndex(String oldStr, int index, char newChar) {
return oldStr.substring(0, index) +
newChar + oldStr.substring(index + 1, oldStr.length());
}
//生成x位的A-Z的字串
public static void doIt(int n) {
int idx = 0;
String start = "";
String[] arr = new String[26];
StringBuffer sb = new StringBuffer(2048);

for (int i = 0; i < n; i++) {
start = start + "A";
}

for (char c = 'A'; c <= 'Z'; c++) {
arr[idx] = replaceIndex(start, n - 1, c);
sb.append(arr[idx]);
sb.append(" ");
idx++;
}
sb.append("\n");
idx = 0;
n--;
while ( (--n) >= 0) {
for (char c = 'B'; c <= 'Z'; c++) {
for (int i = 0; i < 26; i++) {
String tmp = replaceIndex(arr[i], n, c);
sb.append(tmp);
sb.append(" ");
if (c == 'Z') {
arr[i] = tmp;
}
idx++;
}
sb.append("\n");
}
}
System.out.println(sb.toString());
}

public static void main(String args[]) {
doIt(4);
}

这个不知道是不是你要的结果？这个效率应该还算可以。