见《java字符串解析》，联想

JavaEye问答： java字符串解析

本来想回那帖的，但从学校坐了趟公交回到住处那问题就已经关闭了 T T
其实我是想说，这问题用正则表达式解决是如此的简单……因为那帖给的字符串是有规律的： 带有相同数字的项是相邻的。

例如说用Ruby的话，在irb里测试一下：

irb(main):001:0> src = '22:w,22:c,21:r,7:r,7:w,7:c,8:w,8:c,8:d,9:r,9:w,9:c,9:d,10:r'
=> "22:w,22:c,21:r,7:r,7:w,7:c,8:w,8:c,8:d,9:r,9:w,9:c,9:d,10:r"
irb(main):002:0> re = /((\d+):\w(,\2:\w)*),/
=> /((\d+):\w(,\2:\w)*),/
irb(main):003:0> expected_result = '22:w,22:c;21:r;7:r,7:w,7:c;8:w,8:c,8:d;9:r,9:w,9:c,9:d;10:r'
irb(main):004:0> result = src.gsub re, '\1;'
=> "22:w,22:c;21:r;7:r,7:w,7:c;8:w,8:c,8:d;9:r,9:w,9:c,9:d;10:r"
irb(main):005:0> result == expected_result
=> true

也就是说那个问题用一行代码的函数就能解决：

def reformat(src)
  src.gsub /((\d+):\w(,\2:\w)*),/, '\1;'
end

即便用Java也仍然是同样的简单：

public class TestRegex {
    public static String reformat(String src) {
        return src.replaceAll("((\\d+):\\w(,\\2:\\w)*),", "$1;");
    }
    
    public static void main(String[] args) {
        String src = "22:w,22:c,21:r,7:r,7:w,7:c,8:w,8:c,8:d,9:r,9:w,9:c,9:d,10:r";
        String expected = "22:w,22:c;21:r;7:r,7:w,7:c;8:w,8:c,8:d;9:r,9:w,9:c,9:d;10:r";
        String result = reformat(src);
        System.out.println(result.equals(expected));
    }
}

当然如果输入的字符串没有上述的特殊属性就无法保证这个方法的正确性了。只不过我猜那个问题原本作为源的字符串就是从什么别的程序输出得到的，或许本来就有这样的规律，值得利用。