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從逗號分隔整數的字符串數組映射的連續天數

Java

慕尼黑8549860 2023-10-19 21:08:39

我收到了多個以逗號分隔的字符串整數作為輸入，例如以下字符串：“5,6,0”“0,1,2”“1,2,3,4”這些整數中的每一個都代表一周中的一天0 = 星期日 1 = 星期一 2 = 星期二 3 = 星期三 4 = 星期四 5 = 星期五 6 = 星期六對于第一個字符串，這意味著星期四到星期日第二個字符串的有效期是星期日到星期二第三個字符串的有效期是星期一到星期四目前，我正在使用以下 private fun mapOfDays(validDays: String): LinkedHashMap<Int, String> { if (!validDays.isBlank()) { val daysArray = validDays.split("\\s*,\\s*") as Array<String> var mapDays = LinkedHashMap<Int, String>() var mapDay = LinkedHashMap<Int, String>() mapDays[0] = "SUNDAY" mapDays[1] = "MONDAY" mapDays[2] = "TUESDAY" mapDays[3] = "WEDNESDAY" mapDays[4] = "THURSDAY" mapDays[5] = "FRIDAY" mapDays[6] = "SATURDAY" for (day in daysArray) { if (mapDays.containsKey(day.toInt())) { mapDay[day.toInt()] = mapDays[day.toInt()]!! } } return mapDay } return LinkedHashMap() } private fun mappedDays(mapOfDays: LinkedHashMap<Int, String>?): String { if (!mapOfDays.isNullOrEmpty()) { val mapSize = mapOfDays.size if (mapSize > 6) return "All Day" if (mapSize > 5) return sixDayString(mapOfDays) if (mapSize > 4) return fiveDayString(mapOfDays) if (mapSize > 3) return fourDayString(mapOfDays) if (mapSize > 2) return threeDayString(mapOfDays) if (mapSize > 1) return twoDayString(mapOfDays) if (mapSize > 0) return oneDayString(mapOfDays) } return "" }但是，我當前的實現能夠告訴我包含哪些天，但無法繪制出天數組，例如：如果我得到“0,1,3,4,5,6”我想要的最終字符串輸出如下：星期三到星期一或者“0,1,3,4,5”將導致以下結果：Sunday , Monday , Wednesday to Friday。

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2 回答

Qyouu

TA貢獻1786條經驗獲得超11個贊

package days

import java.lang.IllegalArgumentException

class DaysFactory {

fun dayFromInt(index: Int): Day {

return when (index) {

0 -> Day.Sunday

1 -> Day.Monday

2 -> Day.Tuesday

3 -> Day.Wednesday

4 -> Day.Thursday

5 -> Day.Friday

6 -> Day.Saturday

else -> throw IllegalArgumentException("illigal index :$index")

}

enum class Day(val index: Int) {

Sunday(0), Monday(1), Tuesday(2), Wednesday(3), Thursday(4), Friday(5), Saturday(6)

}

class DaysRange(val seed: String) {

var stringFormat = ""

private fun getTomorrow(dayIndex: Int): Int {

if (dayIndex != 6) return dayIndex + 1

return 0

}

override fun toString(): String =stringFormat

init {

if (isValidInput(seed)) {

val dayFactory = DaysFactory()

val indexes = seed.split(",").map { it.toInt() }

val days = indexes.map { dayFactory.dayFromInt(it) }

val ranges = splitIndexesToRanges(indexes)

ranges.forEach { range ->

if (range.size > 2) {

stringFormat += "${dayFactory.dayFromInt(range.first())} to ${dayFactory.dayFromInt(range.last())},"

} else

range.forEach {

stringFormat += "${dayFactory.dayFromInt(it)},"

}

stringFormat = stringFormat.dropLast(1)

}

private fun splitIndexesToRanges(daysRange: List<Int>): ArrayList<List<Int>> {

val result = ArrayList<List<Int>>()

val slicePoint = ArrayList<Int>()

for (i in 0 until daysRange.size - 1) {

if (getTomorrow(daysRange[i]) != daysRange[i + 1]) {

slicePoint.add(i)

}

var start = 0

slicePoint.forEach {

result.add(daysRange.slice(start..it))

start = it + 1

}

result.add(daysRange.slice(start until daysRange.size))

return result

}

private fun isValidInput(seed: String): Boolean = true

fun main(args: Array<String>) {

val input = listOf(

"0,1,2,4,5,6",

"5,6,0",

"1,2,3,4"

)

input.forEach {

val dr = DaysRange(it)

println(dr)

}

示例輸出：

周日至周二、周四至周六

周五至周日

周一至周四

反對回復 2023-10-19

哈士奇WWW

TA貢獻1799條經驗獲得超6個贊

如果可以的話，我會堅持使用給定的時間 API（例如，java.time如果您使用 Java 8 或 joda-time 等）。以下解決方案也適用于您的enum，但您需要對其進行一些調整（即DayOfWeek具有getDisplayName并且還允許添加單日并始終獲得下一個連續天）。

我將工作分成 3 個獨立的任務。

將輸入讀入以下列表DayOfWeek：

fun readInput(input : String) : List<DayOfWeek> = input.splitToSequence(",")

.map(String::toInt)

.map {

/* your 0 is Sunday which is 7 for DayOfWeek; rest is the same */

if (it == 0) 7 else it

}

.map(DayOfWeek::of)

.toList()

也許您想添加.distinct().sorted()它或想事先驗證輸入...這取決于您真正想確保什么...

將星期幾轉換為連續天的列表：

fun List<DayOfWeek>.toDayRangeList() : List<DayRange> = fold(mutableListOf<DayRange>()) { consecutiveDaysList, day ->

consecutiveDaysList.apply {

lastOrNull()?.takeIf { it.to + 1 == day }?.apply {

to = day

} ?: add(DayRange(day))

}

為此，我還引入了一個DateRange-class 以便輕松改變結束日期...您也可以使用不可變對象執行此操作，但我發現這種方式更容易。它還DateRange包括一些輔助方法，可以輕松獲取所需形式的實際日期（在我的示例中FULL_STANDALONE）：

data class DayRange(var from: DayOfWeek, var to: DayOfWeek = from) {

private fun DayOfWeek.toFullString(locale : Locale) = getDisplayName(TextStyle.FULL_STANDALONE, locale)

fun toString(locale : Locale) : String = when (from) {

// TODO add missing locale specific strings!

to -> from.toFullString(locale)

to + 1 -> "All day"

else -> "${from.toFullString(locale)} to ${to.toFullString(locale)}"

}

// just for convenience we use our custom toString-function:

override fun toString() = toString(Locale.getDefault())

}

可選地“展平”列表，即如果最后一天和第一天是連續的，則將它們合并到一個范圍中。當我們直接處理時，DayOfWeek我們可以簡單地添加另一天并比較兩天，無論其中一天是否是一周的最后一天：

fun List<DayRange>.flatten(): List<DayRange> {

if (size > 1) {

val first = first()

val last = last()

if (last.to + 1 == first.from)

return dropLast(1).drop(1)

.toMutableList()

.apply {

add(DayRange(last.from, first.to))

}

return this

}

將它們放在一起/演示：

listOf("1", "1,2", "1,0", "1,2,3", "1,2,4,5", "1,2,4,5,0", "1,2,3,4,5,6,0", "2,3,4,5,6,0,1")

.forEach { input ->

print(input)

readInput(input)

.toDayRangeList()

.flatten()

.joinToString(", ")

.also {

println("-> $it")

}

打印以下內容：

1 -> Monday

1,2 -> Monday to Tuesday

1,0 -> Sunday to Monday

1,2,3 -> Monday to Wednesday

1,2,4,5 -> Monday to Tuesday, Thursday to Friday

1,2,4,5,0 -> Thursday to Friday, Sunday to Tuesday

1,2,3,4,5,6,0 -> All day

2,3,4,5,6,0,1 -> All day

反對回復 2023-10-19

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從逗號分隔整數的字符串數組映射的連續天數

從逗號分隔整數的字符串數組映射的連續天數

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