我有這個(gè) ArrayListArrayList<Character> wrongLetters;還有這個(gè)系統(tǒng)System.out.println("Number of errors: " + wrongLetters.size() + " (" + String.join("", String.valueOf(wrongLetters)) + ")");它現(xiàn)在像這樣打印Number of errors: 9 ([X, M, S, K, B, Q, L, I, U])但我怎樣才能做到這樣Number of errors: 9 (XMSKBQLIU)
3 回答
皈依舞
TA貢獻(xiàn)1851條經(jīng)驗(yàn) 獲得超3個(gè)贊
Java 8 或更高版本：
final String result = wrongLetters.stream().map(String::valueOf).collect(Collectors.joining());
System.out.println("Number of errors: " + wrongLetters.size() + " (" + result + ")");
弒天下
TA貢獻(xiàn)1818條經(jīng)驗(yàn) 獲得超8個(gè)贊
從字符的數(shù)組列表中創(chuàng)建一個(gè)字符串
String str = "";
for (Character c : wrongLetters) {
str += c;
}
然后 ：
System.out.println("Number of errors: " + wrongLetters.size() + " (" + String.join("", str) + ")");
或制作一個(gè)字符串生成器：
StringBuilder stringBuilder= new StringBuilder(wrongLetters.size());
for (Character c : wrongLetters) {
stringBuilder.append(c);
}
然后：
System.out.println("Number of errors: " + wrongLetters.size() + " (" + String.join("", stringBuilder.toString()) + ")");
12345678_0001
TA貢獻(xiàn)1802條經(jīng)驗(yàn) 獲得超5個(gè)贊
愚蠢的答案（只需刪除非字母數(shù)字）：
System.out.println("Number of errors: " + wrongLetters.size() + " (" + String.join("", String.valueOf(wrongLetters).replaceAll("[^A-Za-z0-9]", ""))+ ")");
Java 8+ 的另一個(gè)：
StringJoiner joiner = new StringJoiner("");
wrongLetters.stream().forEach(err -> joiner.add(String.valueOf(err)));
System.out.println("Number of errors: " + wrongLetters.size() + " (" + joiner + ")");
沒(méi)有 Java 8（使用 StringBuilder）：
StringBuilder sb = new StringBuilder(" (");
for (Character character : wrongLetters) {
sb.append(character);
}
sb.append(")");
System.out.println("Number of errors: " + wrongLetters.size() + sb.toString());
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