我有電話號(hào)碼列表，我需要根據(jù)電話類型進(jìn)行過(guò)濾。最終結(jié)果應(yīng)該是單個(gè)對(duì)象 (PhoneUsageType)。示例場(chǎng)景 1. 如果列表有 2 種不同類型的電話號(hào)碼，如 Cell_Phone 和 Home_Phone，它應(yīng)該只返回 Cell_Phone 對(duì)象。2.如果它只有 Cell_Phone 那么它應(yīng)該直接返回我那個(gè)對(duì)象 3.如果它沒(méi)有 Cell_Phone 那么它應(yīng)該返回我 Home_Phone如果 Cell_Phone 在列表中可用，則它應(yīng)該是首選對(duì)象使用list.stream().filter代碼PhoneUsageType phone1 = new PhoneUsageType();PhoneUsageKeyGrpType keyGroup1 = new PhoneUsageKeyGrpType();keyGroup1.setPurposeCode("CELL_PHONE");phone1.setPhoneUsageKeyGrp(keyGroup1);PhoneNumberType phoneNumber1 = new PhoneNumberType();phoneNumber1.setFullNumber("123");phone1.setPhoneNumber(phoneNumber1);PhoneUsageType phone2 = new PhoneUsageType();PhoneUsageKeyGrpType keyGroup2 = new PhoneUsageKeyGrpType();keyGroup2.setPurposeCode("CELL_PHONE");phone2.setPhoneUsageKeyGrp(keyGroup2);PhoneNumberType phoneNumber2 = new PhoneNumberType();phoneNumber2.setFullNumber("456");phone2.setPhoneNumber(phoneNumber2);List<PhoneUsageType> phoneUsageList = Lists.newArrayList();請(qǐng)幫我寫(xiě)下邏輯。提前致謝。如果您需要更多詳細(xì)信息，請(qǐng)告訴我。
2 回答
隔江千里
TA貢獻(xiàn)1906條經(jīng)驗(yàn) 獲得超10個(gè)贊
不要按類型使用filter、使用min或排序。max
像下面這樣的東西：
enum PhoneType {
CELL, WORK, HOME; // in order of preference
}
class Phone {
public PhoneType getType() ...
public String getNumber() ...
}
Phone best = Collections.min(phoneList, Comparator.comparing(Phone::getType));
梵蒂岡之花
TA貢獻(xiàn)1900條經(jīng)驗(yàn) 獲得超5個(gè)贊
更好地使用最小和最大排序類型
enum PhoneType {
Mobile, Office, personal; // in order of preference
}
class Mobile {
public PhoneType getType()
public String getNumber()
}
Comparator<Mobile> preferredNumber = Comparator.comparing(Mobile::getType);
Mobile best = mobileList.stream().min(preferredNumber).orElseThrow();
希望你能明白。
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