Question

我正在尝试计算单词的拼字游戏得分。但是，每次我输入一个单词时，它都会返回每个字母与之关联的数字序列。例如，当我输入fox作为我拼字游戏的单词时，我得到418。为什么会这样？

const oldScoreKey = {
  1: ['A', 'E', 'I', 'O', 'U', 'L', 'N', 'R', 'S', 'T'],
  2: ['D', 'G'],
  3: ['B', 'C', 'M', 'P'],
  4: ['F', 'H', 'V', 'W', 'Y'],
  5: ['K'],
  8: ['J', 'X'],
  10: ['Q', 'Z']
};


function transform(oldScoreKey){
  const newScoreKey = {};
  for (const [letterValue, letterArr] of Object.entries(oldScoreKey)) {
    for (const letter of letterArr) {
      newScoreKey[letter.toLowerCase()] = letterValue;
    }
  }
  return newScoreKey;
}

console.log(transform(oldScoreKey));


// Code your initialPrompt function here:
const input = require('readline-sync');

console.log("Using algorithm: Scrabble");

let word = (input.question("Enter a simple word please: "));
const alphabet = "abcdefghijklmnopqrstuvwxyz";


function simpleScore() {
  let score = 0

  for(let letter of word.toLowerCase()){
    if (alphabet.includes(letter))
    score += 1;
  }
  return score;
}

console.log(simpleScore())


let letter = (input.question("Enter a scrabble word please: "));
letter = letter.toLowerCase();


let newAlphabet = { a: '1',
  e: '1',
  i: '1',
  o: '1',
  u: '1',
  l: '1',
  n: '1',
  r: '1',
  s: '1',
  t: '1',
  d: '2',
  g: '2',
  b: '3',
  c: '3',
  m: '3',
  p: '3',
  f: '4',
  h: '4',
  v: '4',
  w: '4',
  y: '4',
  k: '5',
  j: '8',
  x: '8',
  q: '10',
  z: '10' }
function scrabbleScore() {
let sum = 0
let i = 0
let score = 0
for (i = 0; i < word.length; i++) {
    letter = word[i];
    sum += newAlphabet[letter];
}
return (sum*1);
}
console.log(scrabbleScore())

打印Scrabble分数时，我试图获得总分418，这等于13分。

Answer 1

分数应为数字。这是表示此操作的简单方法。

[
  {
    "genre_id": "",
    "name": "",
    "description": "",
    "slug": "",
    "url": "",
    "videos": [
       {
         "videos_id": "",
         "title": "",
        "description": "",
        "slug": "",
        "release": "",
        "is_tvseries": "",
        "runtime": "",
        "video_quality": "",
        "thumbnail_url": "",
        "poster_url": ""
      },
      {
        "videos_id": "",
        "title": "",
        "description": "",
        "slug": "",
        "release": "",
        "is_tvseries": "",
        "runtime": " ",
        "video_quality": "",
        "thumbnail_url": "",
        "poster_url": ""
      }
    ]
    }, 
    {
    "genre_id": "",
    "name": "",
    "description": " ",
    "slug": "",
    "url": "",
    "videos": [
      {
        "videos_id": "",
        "title": " ",
        "description": "",
        "slug": "",
        "release": "",
        "is_tvseries": "",
        "runtime": "",
        "video_quality": "",
        "thumbnail_url": "",
        "poster_url": ""
      },
      {
        "videos_id": "",
        "title": "",
        "description": "",
        "slug": "",
        "release": "",
        "is_tvseries": "",
        "runtime": "",
        "video_quality": "",
        "thumbnail_url": "",
        "poster_url": ""
      } 
    ] 
  } 
]

Answer 2

因为它不会转换为整数。我将从您的代码中提取代码。

function scrabbleScore() {
    let sum = 0;
    let i = 0;
    let score = 0;
    for (i = 0; i < word.length; i++) {
        letter = word[i];
        sum += parseInt(newAlphabet[letter]);
    }
    return (sum*1);
}
console.log(scrabbleScore());

这将为您提供帮助。

Answer 3

如果对象中不存在字符，则可以使用数字和默认值为零的对象。

function scrabbleScore(word) {
    let newAlphabet = { a: 1, e: 1, i: 1, o: 1, u: 1, l: 1, n: 1, r: 1, s: 1, t: 1, d: 2, g: 2, b: 3, c: 3, m: 3, p: 3, f: 4, h: 4, v: 4, w: 4, y: 4, k: 5, j: 8, x: 8, q: 10, z: 10 },
        sum = 0,
        i;

    word = word.toLowerCase();
    for (i = 0; i < word.length; i++) {
        sum += newAlphabet[word[i]] || 0; // for unknown characters
    }
    return sum;
}

let letter = prompt("Enter a scrabble word please: ");

console.log(scrabbleScore(letter))

Answer 4

在JS中，+=对于字符串来说是可以的，它不将值解析为int而是仅进行连接。尝试手动强制：

sum += parseInt(newAlphabet[letter]);

Answer 5

这是我要分解的方式。

此版本包括用于创建新得分密钥的代码的较干净版本。（取决于flatMap尚不通用，但易于填充，并且可以根据需要用reduce轻松替换。）

const invertScoreKey = (old) =>
  Object.assign (...Object .entries (old) .flatMap (
    ([k, arr]) => arr .map ((v) => ({[v .toLowerCase ()]: Number (k)}))
  ))

const sum = (ns) => ns .reduce((total, n) => total + n, 0)

const scoreWord = (scores) => (word) => 
  sum ([...word .toLowerCase()] .map(c => scores[c] || 0))

const oldScoreKey = {
  1: ['A', 'E', 'I', 'O', 'U', 'L', 'N', 'R', 'S', 'T'],
  2: ['D', 'G'],
  3: ['B', 'C', 'M', 'P'],
  4: ['F', 'H', 'V', 'W', 'Y'],
  5: ['K'],
  8: ['J', 'X'],
  10: ['Q', 'Z']
}

const scrabbleScore = scoreWord (invertScoreKey (oldScoreKey))

'The quick, brown fox jumped over the lazy dog'.split(/\W+/)
  .forEach(word => console.log(`${word} : ${scrabbleScore(word)}`)
)

请注意，scoreWord是通用的，接受一个分数键并返回一个函数，该函数接受一个单词并通过该键返回总分数。 scrabbleScore以此为基础，使用了将原始密钥反转为密钥的结果。

计算拼字游戏Java脚本的得分

5 个答案: