我需要编写一段代码,它接受一个输入字符串参数,并确定每对加起来为 10 的数字之间是否正好存在 3 个问号。如果是,则返回 true,否则返回 false。一些示例测试用例在下面
"arrb6???4xxbl5???eee5" => true
"acc?7??sss?3rr1??????5" => true
"5??aaaaaaaaaaaaaaaaaaa?5?5" => false
"9???1???9???1???9" => true
"aa6?9" => false
我已经尝试在 Java 中实现它,如下所示,但结果并不如预期
public static String QuestionsMarks(String str) {
str = str.replaceAll("[a-z]+","");
Pattern pattern = Pattern.compile("([0-9])([?])([?])([0-9])");
Pattern pattern01 = Pattern.compile("([0-9])([?])([?])([0-9])");
Matcher matcher01 = pattern01.matcher(str);
Pattern pattern02 = Pattern.compile("([0-9])([0-9])");
Matcher matcher02 = pattern02.matcher(str);
Matcher matcher = pattern.matcher(str);
if (matcher01.find() || matcher02.find()) {
return "false";
} else if (matcher.find()) {
return "true";
}
return "false";
}
答案 0 :(得分:2)
检测用多个问号分隔的重叠数字的正则表达式如下:
(?=((\d)([^?\d]*\?[^?\d]*)*(\d)))
它使用正向前瞻 (?=)
来处理此模式的相邻匹配项之间的重叠数字:
(\d)([^?\d]*\?[^?\d]*)*(\d)
:由任意数量的问号和其他可选字符分隔的两位数字。
因此,正则表达式提供了以下形式的匹配流:Group1=StringWithQuestionMarks, Group2=Digit, Group4=Digit
,它应该根据规则进行验证:将数字总和计算为 10,将问号之间的问号计算为 3。< /p>
使用 Stream API 的示例实现:
private static final Pattern DIGS = Pattern.compile("(?=((\\d)([^?\\d]*\\?[^?\\d]*)*(\\d)))");
public static boolean hasTotal10Around3QMarks(String str) {
Supplier<Stream<MatchResult>> ss = () -> DIGS
.matcher(str)
.results()
.filter(r -> r.groupCount() == 4
&& 10 == Integer.parseInt(r.group(2)) + Integer.parseInt(r.group(4))
);
return ss.get().findAny().isPresent() &&
ss.get().allMatch(r -> 3 == r.group(1).chars()
.filter(c -> c == '?')
.count()
);
}
Supplier<Stream>
有助于重用 MatchResult
流来处理最后一个测试用例,当没有匹配的数字对且中间有 3 个问号时 (allMatch
为空流返回 {{1} }).
测试:
true
输出:
String[] tests = {
"arrb6???4xxbl5???eee5",
"acc?7??sss?3rr1??????5",
"5??aaaaaaaaaaaaaaaaaaa?5?5",
"9???1???9???1???9",
"aa6?9",
""
};
Arrays.stream(tests)
.forEach(t -> System.out.printf("'%s' => %s%n", t, hasTotal10Around3QMarks(t)));