5

我需要生成递增的字符序列,其中每个字符可以是不同的字母或数字范围。有谁知道有这样一个任务的图书馆?例如:

AAA000_A0

其中 A 是任意字母 AZ,0 是任意数字 0-9。我也需要增加它们,例如:

AAA000_A0++ = AAA000_A1

如果你继续前进,它会像这样延续:

AAA000_A9++ = AAA000_B0

直到达到:

ZZZ999_Z9
4

2 回答 2

3

我会将问题分解到每个点,有点像汽车的里程表:

public class CounterDigit {
    private final CounterDigit _higher;
    private final String _chars;
    private int _index = 0;

    public CounterDigit(CounterDigit higher, String chars) {
        _higher = higher;
        _chars = chars;
    }

    public CounterDigit(CounterDigit higher, char first, char last) {
        this(higher, charRange(first, last));
    }

    private static String charRange(char first, char last) {
        StringBuilder b = new StringBuilder();
        for (char c = first; c <= last; ++c) {
            b.append(c);
        }
        return b.toString();
    }

    public char current() {
        return _chars.charAt(_index);
    }

    public void increment() {
        if (++_index >= _chars.length()) {
            _index = 0;
            if (_higher != null) {
                _higher.increment();
            }
        }
    }
}

然后,您可以聚合这些的有序序列:

public class Counter {
    private CounterDigit[] _digits;

    public Counter(String template) {
        CounterDigit recent = null;
        _digits = new CounterDigit[template.length()];
        for (int i = 0; i < template.length(); ++i) {
            char c = template.charAt(i);
            if      (c == 'A') { recent = new CounterDigit(recent, 'A', 'Z'); }
            else if (c == '0') { recent = new CounterDigit(recent, '0', '9'); }
            else               { recent = new CounterDigit(recent, c, c);     }
            _digits[i] = recent;
        }
    }

    public void increment() {
        _digits[_digits.length-1].increment();
    }

    @Override
    public String toString() {
        StringBuilder b = new StringBuilder(_digits.length);
        for (CounterDigit digit : _digits) {
            b.append(digit.current());
        }
        return b.toString();
    }
}
于 2013-05-29T20:06:03.383 回答
1

这是一个简单的算法,一个简短的函数就足够了一个库。

这是我放在一起的 JavaScript 示例。

var _LETTERS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
var _NUMBERS = "0123456789";
function nextValue(str) {
    var result = "";
    var carry = true;
    var chr, idx, lookup;
    for(var i=str.length - 1; i>=0; i--) {
        chr = str.charAt(i);
        if(carry) {
            lookup = _LETTERS.indexOf(chr) >= 0 ? _LETTERS : (_NUMBERS.indexOf(chr) >= 0 ? _NUMBERS : "");
            idx = lookup.indexOf(chr);
            if(idx >= 0) {
                if(idx == lookup.length - 1) {
                    idx = 0;
                } else {
                    idx++;
                    carry = false;
                }
                chr = lookup.charAt(idx);
            }
        }
        result = chr + result;
    }
    return result;
}

var demo = "0_A";
while(demo != "9_Z") {
    demo = nextValue(demo);
    console.log(demo);
}

移植/转换为 Java 应该很简单,“indexOf”和“charAt”等函数名称应该相同。

于 2013-05-29T19:41:45.510 回答