Question

我认为,这似乎有局限性......这相当于我sat倒并开始打字。

不管怎么说,我的任务是撰写一种方法,在“强”中找到“主角”,并适当取代它们,从而产生结果。职能声明将考虑如下内容:

public String doStuff(String source, Map<Integer, String> replacementData)

想法是,每一“标签”都将是四舍五入的“配对、加固、加固”和“特”类(因此,切除是“......<<[0-9]+>>)。一旦在<代码>源上碰到这些标签之一,标签内所含的立体一就应当从标签中提取,标记应当由<代码>replacementData.get(I)替换。

无论如何,我要有效地这样做。我不敢肯定如何做到这一点。我对“模式”和“匹配”进行了研究,我认为,这是正确的路线。我倾向于不使用Sting.index。这与这一点一样,因为其效率不高(即没有效率)。

感谢大家的建议!

Answer 1

The way to do it with Regex Pattern and Matcher is to capture the integer in what is known as a group and then use that to replace. The other interesting part of your problem is the step by step replacement. The sequence of steps that you need to do are described in the javadoc for appendReplacement method

例:

String str = "abc<<1>>def<<2>>ghi";
java.util.Map<Integer,String> replacementMap = new java.util.HashMap<Integer,String>() {
    {
        put(1," ONE ");
        put(2," TWO ");
    }
};
java.util.regex.Pattern pattern = java.util.regex.Pattern.compile("<<([0-9]+)>>");
java.util.regex.Matcher matcher = pattern.matcher(str);
int start = 0;
StringBuffer sb = new StringBuffer();
while(matcher.find()) {
    matcher.appendReplacement(sb,replacementMap.get(Integer.parseInt(matcher.group(1))));
}
matcher.appendTail(sb);
System.out.println(sb.toString());

产出:abc ONE def TWO ghi

Answer 2

执行:

public String doStuff(String source, Map<Integer, String> replacementData){

    for( Map.Entry<Integer,String> entry : replacementData.entrySet() )
         source = source.replace("<<"+entry.getKey().toString()+">>", entry.getValue() );

    return source;
}

~~这对我来说,没有太低的效率。提高其效率的唯一途径是设法在座标上填满一封信件,但这实际上需要改写<代码>String.replace(.~~。

~~考虑到将投入的工作量,我只想把这种优化归入某类工作,只有当简介者表明这是真正的瓶颈。~~

The Matcher-based method by Puneet is better since it only passes over the string once.

友情链接