这里有3个解决方案：前两个是算术增量，而第三个更多是字符操作。

这三个实现都通过相同的单元测试：

 assertEquals("1DDA01A", MyClass.increment("1DDA00Z"));
 assertEquals("1A9AV00", MyClass.increment("1A9AU99"));
 assertEquals("AFH00", MyClass.increment("AFG99"));
 assertEquals("A2GF24", MyClass.increment("A2GF23"));
 assertEquals("ABAA0000", MyClass.increment("AAZZ9999"));
 assertEquals("11AB0A", MyClass.increment("11AA9Z"));

第一：

public static String increment(String number) {
    Pattern compile = Pattern.compile("^(.*?)([9Z]*)$");
    Matcher matcher = compile.matcher(number);
    String left="";
    String right="";
    if(matcher.matches()){
         left = matcher.group(1);
         right = matcher.group(2);
    }
    number = !left.isEmpty() ? Long.toString(Long.parseLong(left, 36) + 1,36):"";
    number += right.replace("Z", "A").replace("9", "0");
    return number.toUpperCase();
}

第二：

public static String increment(String number) {
    Pattern compile = Pattern.compile("^(.*?)([0-9]*|[A-Z]*)$");
    Matcher matcher = compile.matcher(number);
    String remaining = number;
    String currentGroup = "";
    String result = "";

    boolean continueToNext = true;
        while (matcher.matches() && continueToNext) {
        remaining = matcher.group(1);
        currentGroup = matcher.group(2);
        int currentGroupLength = currentGroup.length();
        int base = currentGroup.matches("[0-9]*") ? 10 : 36;
        currentGroup = Long.toString(Long.parseLong("1" + currentGroup, base) + 1, base);  // The "1" if just to ensure that "000" doesn't become 0 (and thus losing the original string length)
        currentGroup = currentGroup.substring(currentGroup.length() - currentGroupLength, currentGroup.length());
        continueToNext = Long.valueOf(currentGroup, base) == 0;
        if (base == 36) {
            currentGroup = currentGroup.replace("0", "A");
        }

        result = currentGroup + result;
        matcher = compile.matcher(remaining);
    }

    result = remaining + result;
    return result.toUpperCase();
}

第三 ：

这适用于您当前的“要求”。与开始时所问的问题相比，这不仅仅是“由字母组成的左半部分”
+“由数字组成的右半部分”。现在，“一切都变了”，字母从A滚动到Z到A，数字从0滚动到9到0。当字母到达Z时，它将重置为A，然后左侧的数字/字母递增。

如果所有数字均递增，则不会在左侧添加新数字。您没有在问题中提到这一点，但是我敢肯定您可以从这里弄清楚这一点：

public static String increment(String number) {
    char[] cars = number.toUpperCase().toCharArray();
    for (int i = cars.length - 1; i >= 0; i--) {
        if (cars[i] == 'Z') {
            cars[i] = 'A';
        } else if (cars[i] == '9') {
            cars[i] = '0';
        } else {
            cars[i]++;
            break;
        }
    }
    return String.valueOf(cars);
}

至于“计数”，您的示例不足以理解逻辑。它仅计算数字吗？字母呢？它遵循baseXx吗？

AA010-AAA003 = 7，3 A与2 A无关吗？我觉得这是您要了解您的要求（即：家庭作业..）

从技术上讲，这可以回答最初提出的问题（在此过程中进行了许多修改）。