這里的代碼基本上用于根據(jù)骰子的滾動(dòng)來決定字符串。其中有很多陳述，我確實(shí)需要一些幫助來減少這種情況。我在網(wǎng)上看了一些方法，但不太適合。public String generatePassageSection(){ int roll = roll(); if(roll<=2 && roll>=1) { return "passage goes straight for 10 feet."; } else if(roll<=5 && roll>=3) { return "passage ends in door to a chamber."; } else if(roll<=7 && roll>=6) { return "door to right (main passage continues straight for 10 ft)"; } else if(roll<=9 && roll>=8) { return "door to left (main passage continues straight for 10 ft)"; } else if(roll<=11 && roll>=10) { return "passage turns to left and continues for 10 ft"; } else if(roll<=13 && roll>=12) { return "passage turns to right and continues for 10 ft"; } else if(roll<=16 && roll>=14) { return "passage ends in door to chamber"; } else if(roll==17) { return "Stairs, (passage continues straight for 10 ft)"; } else if (roll<=19 && roll>=18) { return "Dead end"; } else if(roll==20) { return "Wandering Monster (passage continues straight for 10 ft)"; } else { return null; }
3 回答
大話西游666
TA貢獻(xiàn)1817條經(jīng)驗(yàn) 獲得超14個(gè)贊
這是一個(gè)非常好的問題。因?yàn)槊總€(gè) if-else 都是一樣的；您檢查某個(gè)滾動(dòng)是否在某個(gè)范圍內(nèi)，然后返回一個(gè)字符串，您可以只使用枚舉。相對(duì)于您正在處理和用此數(shù)據(jù)表示的內(nèi)容，可以改進(jìn)枚舉名稱以及元素名稱。
public String generatePassageSection() {
int roll = roll();
Message message = Stream.of(Message.values()).filter(m -> m.inBounds(roll)).findAny().orElse(null);
return message == null ? null : message.message;
}
enum Message {
FIRST(1, 2, "passage goes straight for 10 feed."),
SECOND(3, 5, "passage ends in the door to a chamber."),
THIRD(6, 7, "door to right (main passage continues straight for 10 ft)"),
FOURTH(9, 10, "door to left (main passage continues straight for 10 ft)"),
FIFTH(10, 11, "passage turns to left and continues for 10 ft"),
SIXTH(12, 13, "passage turns to right and continues for 10 ft"),
SEVENTH(14, 16, "passage ends in door to chamber"),
EIGHTH(17, 17, "Stairs, (passage continues straight for 10 ft)"),
NINTH(18, 19, "Dead end"),
TENTH(20, 20, "Wandering Monster (passage continues straight for 10 ft)");
;
private final int minimumRollInclusive;
private final int maximumRollInclusive;
private final String message;
Message(int minimumRollInclusive, int maximumRollInclusive, String message) {
this.minimumRollInclusive = minimumRollInclusive;
this.maximumRollInclusive = maximumRollInclusive;
this.message = message;
}
boolean inBounds(int roll) {
return roll >= minimumRollInclusive && roll <= maximumRollInclusive;
}
}
int roll() {
return 0; // use ur code
}
另外，應(yīng)該注意的是，Enum#values 每次調(diào)用時(shí)都會(huì)創(chuàng)建一個(gè)新的數(shù)組對(duì)象，因此值得對(duì)其進(jìn)行緩存。
郎朗坤
TA貢獻(xiàn)1921條經(jīng)驗(yàn) 獲得超9個(gè)贊
如果您只是想減少 if 語句的數(shù)量，您可以添加一組可能的結(jié)果，在適當(dāng)?shù)镊蛔由蠈?duì)每個(gè)結(jié)果進(jìn)行索引，然后返回該索引處的值。我不會(huì)把它全部寫給你，但它看起來有點(diǎn)像這樣：
public String generatePassageSection(String[] diceRolls){
return diceRolls[roll()];
}
子衿沉夜
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超3個(gè)贊
更喜歡數(shù)據(jù)結(jié)構(gòu)而不是代碼。
final private static String text[] = {
"passage goes straight for 10 feet.", // 1, 2
"passage ends in door to a chamber.", // 3, 4, 5
"door to right (main passage continues straight for 10 ft)", // 6, 7
"door to left (main passage continues straight for 10 ft)", // 8, 9
"passage turns to left and continues for 10 ft", // 10, 11
"passage turns to right and continues for 10 ft", // 12, 13
"passage ends in door to chamber", // 14, 15, 16
"Stairs, (passage continues straight for 10 ft)", // 17
"Dead end", // 18, 19
"Wandering Monster (passage continues straight for 10 ft)" // 20
}
final private static int index[] = {
0, 0, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 8, 8, 9
}
public String generatePassageSection() {
return text[index[roll()-1]];
}
手動(dòng)構(gòu)造的索引數(shù)組對(duì)文本數(shù)組的順序和內(nèi)容有隱式假設(shè)。就本次例行的情況和規(guī)模而言，我認(rèn)為這是合理的；我不建議將其作為一般做法。
這隱含地假設(shè) roll() 已知能夠可靠地返回 1 到 20（含）范圍內(nèi)的結(jié)果；如果你不能信任它，應(yīng)該添加錯(cuò)誤檢查。
添加回答
舉報(bào)
0/150
提交
取消