我的版本实现了Iterator并且保持一个int计数器.计数器值转换为相应的字符串：

import com.google.common.collect.AbstractIterator;

class Sequence extends AbstractIterator {

private int now;

private static char[] vs;

static {

vs = new char['Z' - 'A' + 1];

for(char i='A'; i<='Z';i++) vs[i - 'A'] = i;

}

private StringBuilder alpha(int i){

assert i > 0;

char r = vs[--i % vs.length];

int n = i / vs.length;

return n == 0 ? new StringBuilder().append(r) : alpha(n).append(r);

}

@Override protected String computeNext() {

return alpha(++now).toString();

}

}

在Iterator上调用next()来使用它.

Sequence sequence = new Sequence();

for(int i=0;i<100;i++){

System.out.print(sequence.next() + " ");

}

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 AA AB AC AD AE

较大序列具有更好性能的实现会重复使用公共前缀：

class SequencePrefix extends AbstractIterator {

private int now = -1;

private String prefix = "";

private static char[] vs;

static {

vs = new char['Z' - 'A' + 1];

for(char i='A'; i<='Z';i++) vs[i - 'A'] = i;

}

private String fixPrefix(String prefix){

if(prefix.length() == 0) return Character.toString(vs[0]);

int last = prefix.length() - 1;

char next = (char) (prefix.charAt(last) + 1);

String sprefix = prefix.substring(0, last);

return next - vs[0] == vs.length ?

fixPrefix(sprefix) + vs[0] : sprefix + next;

}

@Override protected String computeNext() {

if(++now == vs.length) prefix = fixPrefix(prefix);

now %= vs.length;

return new StringBuilder().append(prefix).append(vs[now]).toString();

}

}

如果您使用与数组一起使用的实现来重写此基本算法,那么您将获得更好的性能. (String.charAt,String.substring和StringBuffer有一些开销)