string - 如何在字符串中找到得分最高的单词?
问题描述
我又在做一些 CodeWars 挑战。
我有一个问题:
“给定一串单词,你需要找到得分最高的单词。
单词的每个字母根据其在字母表中的位置得分:a = 1、b = 2、c = 3 等。
您需要将得分最高的单词作为字符串返回。
如果两个单词得分相同,则返回原始字符串中出现最早的单词。
所有字母都是小写的,所有输入都是有效的。”
我已经在 SO 上查找了解决方案,并使用了一个人的想法(尽管我确实改变了一点)。
它仍然不起作用。
有任何想法吗?
object Scoring {
def high(s: String): String = {
var max = 0
var whichWord = 0
var x = 0
var y = new Array[Int](100)
for(word <- s.split(" ")){
for(letter <- word) letter match{
case 'a' => y(x)+=1
case 'b' => y(x)+=2
case 'c' => y(x)+=3
case 'd' => y(x)+=4
case 'e' => y(x)+=5
case 'f' => y(x)+=6
case 'g' => y(x)+=7
case 'h' => y(x)+=8
case 'i' => y(x)+=9
case 'j' => y(x)+=10
case 'k' => y(x)+=11
case 'l' => y(x)+=12
case 'm' => y(x)+=13
case 'n' => y(x)+=14
case 'o' => y(x)+=15
case 'p' => y(x)+=16
case 'q' => y(x)+=17
case 'r' => y(x)+=18
case 's' => y(x)+=19
case 't' => y(x)+=20
case 'u' => y(x)+=21
case 'v' => y(x)+=22
case 'w' => y(x)+=23
case 'x' => y(x)+=24
case 'y' => y(x)+=25
case 'z' => y(x)+=26
}
x +=1
}
for(x <- 0 until y.length){
if(y(x) > max)
{
max = y(x)
whichWord = x
}
}
s.substring(whichWord-1, whichWord)
}
}
以下是测试:
试验结果:
RandomTestSpec
high("ykvhorsqve kfkq jhjibercdptf efevxax ccr vnsmumqby jwhxvamegupfcj lierziuopbcsutm") should return "lierziuopbcsutm"
Test Failed
"[s]" was not equal to "[lierziuopbcsutm]"
Stack Trace
Completed in 34ms
high("skwwwm") should return "skwwwm"
Test Failed
String index out of range: -1
Stack Trace
Completed in 1ms
high("a") should return "a"
Test Failed
String index out of range: -1
Stack Trace
Completed in 1ms
high("gykoialocufuc wcdwuxksqk bvapztcnqwx") should return "bvapztcnqwx"
Test Failed
"[y]" was not equal to "[bvapztcnqwx]"
Stack Trace
Completed in 1ms
high("gdhfoonwtih xbvsiaqhsesl obrndpz nxt inkklyo lf oyoadxqhuys lbqr oxbqq bopalqknjxvpg") should return "oyoadxqhuys"
Test Failed
"o[]" was not equal to "o[yoadxqhuys]"
Stack Trace
Completed in 1ms
high("bbzlmqhsypato pfufsi ryu oboklfa iigha h m") should return "bbzlmqhsypato"
Test Failed
String index out of range: -1
Stack Trace
Completed in 1ms
high("dbtfwvhk kadarmvldjhkx dgxffryldcxodtn hoffibiayxriqe gtqzeuywpgc nqlgvudy") should return "dgxffryldcxodtn"
Test Failed
"[b]" was not equal to "[dgxffryldcxodtn]"
Stack Trace
Completed in 1ms
high("vhyxdefryeznlkz fcaenzsnoxsn phdqu zjbbbybjmdn dbfhvxwnusz dqbqskfbwuomzsl ogsdioilk") should return "vhyxdefryeznlkz"
Test Failed
String index out of range: -1
Stack Trace
high("yqbzfuprmezbgee yxsewucg u") should return "yqbzfuprmezbgee"
Test Failed
String index out of range: -1
Stack Trace
Completed in 1ms
high("zifha") should return "zifha"
Test Failed
String index out of range: -1
Stack Trace
high("moroydyolj tcfpokvitzwzor rnzeacau") should return "tcfpokvitzwzor"
Test Failed
"[m]" was not equal to "[tcfpokvitzwzor]"
Stack Trace
Completed in 1ms
high("jhieih m") should return "jhieih"
Test Failed
String index out of range: -1
Stack Trace
high("yn ounbzw wk eldpjyikbfs nzm") should return "eldpjyikbfs"
Test Failed
"[ ]" was not equal to "[eldpjyikbfs]"
Stack Trace
Completed in 1ms
解决方案
在 Scala 中,使用集合中提供的函数更容易/(更好)。
在您的示例中-这里有一种可能性:
首先要添加得分,您可以使用它:
"hello".map(_.toInt).sum // 532
这会将所有字符的总和作为 Int 值返回 (a=97; ...; z=122)
要找到您可以使用的最高权重foldLeft
,从“”开始。
scala> List("a", "ab","ba").foldLeft("")((a,b)=> higher(a,b))
res10: String = ab
这里完整的higher
功能:
def higher(a:String, b:String):String=
if(a.map(_.toInt).sum >= b.map(_.toInt).sum) a else b
使用集合函数有很多很酷的可能性 -不要使用可变状态!这是切换到 Scala 的一个重要原因。
更新:在https://www.codewars.com上玩后,我发现了一个错误和一个更短的解决方案:
s.split(" ").map(w => (w, w.map(_.toInt - 96).sum)).maxBy(_._2)._1
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