我可以使此代码更高效/更短至运行吗？

Question

所以我对 java 脚本和一般编码还很陌生。我制作一个 Wordle 算法只是为了好玩，以培养我的编码技能。在制作这个算法时，我意识到我将不得不重新创建 Wordle。所以我在网上寻找 java 脚本中的 wordle 娱乐，我发现所有这些我都看到了一个我不想拥有的缺陷。

我看到的缺陷是在 IF 语句中检查 wordle 答案中的字母。特别是当字母在单词中但不在正确的位置时。我看到的游戏有一个 IF 语句，看起来像这样。

IF (answerWord.includes(guessLetter[i])) {
        guessLetter[i] = color.(yellow)
}

（这不是他们写的，但它是主要思想）

我主要关注 .includes。这是行不通的，因为假设我们的猜测词是“同意”而答案是“准备好”。 “同意”中的 2 个 E 是黄色的，我们不希望这样，因为它们在“准备就绪”中只有 1 个 E。所以我们只希望“同意”中的第一个 E 是黄色而不是第二个 E.

所以我决定自己想办法解决这个问题，我想出了这个办法，而且效果很好。至少我很确定它是有效的，基于我测试过的许多词。

所以我的问题是，由于我正在制作一个算法，我将进行大量计算，这是我能写出的最有效的算法还是我可以改进它？

let guessWord = "agree"; // the first word thats inputed
let answerWord = "ready"; // the answer to the wordle

/* 
  'letterCheck' is an array that tells what condition each letter in 'startLetter' is, based on the answer word
    2 = the letter is in the word and in the correct place (GREEN)
    1 = the letter is in the word but not in the correct place (YELLOW)
    0 = the letter in not in the word (GREY)
*/
var letterCheck = ["0", "0", "0", "0", "0"];
var guessLetter = ["A", "A", "A", "A", "A"]; // the separated letters of 'startword' in a array
var answerLetter = ["A", "A", "A", "A", "A"]; // the separated letters of 'answord' in a array

//adds the start word and the answer world to the arrays
for (var i = 0; i < 5; i++) {
  guessLetter[i] = guessWord.substring(i, i + 1);
  answerLetter[i] = answerWord.substring(i, i + 1);
}

console.log(guessLetter);
console.log(letterCheck);
console.log(answerLetter);

//this loops goes though every letter one by one 
for (var i = 0; i < 5; i++) {
  //checks if the letter is in the right spot
  if (guessLetter[i] == answerLetter[i]) {
    letterCheck[i] = "2";
    console.log(guessLetter[i]);
    console.log(letterCheck);
  } else if (answerWord.includes(guessLetter[i])) {
    //checks if there is more than one letter in start word or its the first letter
    if (guessWord.split(guessLetter[i]).length - 1 == 1 || i == 0) {
      letterCheck[i] = "1";
      console.log(guessLetter[i]);
      console.log(letterCheck);
      //checks if the the amount of same letters in start words is equel to the amount of same letters in answer word
    } else if (guessWord.split(guessLetter[i]).length - 1 == answerWord.split(guessLetter[i]).length - 1) {
      letterCheck[i] = "1";
      console.log(guessLetter[i]);
      console.log(letterCheck);
      //opposite of above
    } else if (guessWord.split(guessLetter[i]).length - 1 != answerWord.split(guessLetter[i]).length - 1) {
      letterCheck[i] = "1";
      console.log(guessLetter[i]);
      console.log(letterCheck);
      // checks if any of the letters infront of it are the same as it
      for (var j = 0; j < i; j++) {
        if (guessLetter[i] == guessLetter[j]) {
          letterCheck[i] = "0";
          console.log(guessLetter[i]);
          console.log(letterCheck);
        }
      }
    }
  } else {
    letterCheck[i] = "0";
    console.log(guessLetter[i]);
    console.log(letterCheck);
  }
}
console.log(guessLetter);
console.log(letterCheck);
console.log(answerLetter);

（稍后我将删除 console.log，他们只是帮助我调试。）

抱歉，如果我的代码可能令人困惑，我不是保持代码整洁的最佳人选。如果您有任何困惑或疑问，我会尽力消除任何困惑。

Answer 1

好吧，在简化过程中需要考虑以下事项。您可以使用简单的 MAP 循环来确定哪些字母不正确，然后在单词中突出显示这些字母。

let guessWord = "agree"; // the first word thats inputed
let answerWord = "ready"; // the answer to the wordle

let tmpGuess = guessWord.split('');
let incorrects = answerWord.split('').map(e => {
  let i = tmpGuess.indexOf(e);
  if (i > -1) { // found it!
    tmpGuess.splice(i, 1); // remove
    return '';
  }
  return e;
}).filter(f => f);

console.log(incorrects);

Answer 2

要获得所有状态（按优先顺序：正确，wrong-position，不正确），您应该检查该优先顺序，并在设置该字母位置的状态后将字母从争用中移除.

function check(guess, targetWord) {
    const checkStates = new Array(5).fill("incorrect");
    const letters = guess.split('');
    const targetLetters = targetWord.split('');
    for (let i=0; i<5; i++) {
      if (targetLetters[i] === letters[i]) {
        checkStates[i] = "correct";
        targetLetters[i] = '';
        letters[i] = '';
      }
    }
    for (let i=0; i<5; i++) {
      if (!letters[i]) continue;
      let fi = targetLetters.findIndex(t => t===letters[i]);
      if (fi > -1) {
        checkStates[i] = "contains";
        targetLetters[fi] = '';
      }
    }
    return checkStates;
}

let solution = "ready"
console.log(`solution is ${solution}`);

let guess = "agree";
console.log(`checking ${guess}... ${check(guess, solution)}`);

guess = "roars";
console.log(`checking ${guess}... ${check(guess, solution)}`);

guess = "boars";
console.log(`checking ${guess}... ${check(guess, solution)}`);

solution = "beast"
console.log(`\nsolution is ${solution}`);

guess = "treat";
console.log(`checking ${guess}... ${check(guess, solution)}`);

我可以使此代码更高效/更短至运行吗？

Can i make this code more efficient / shorter to run?

javascript

algorithm

wordle-game

我可以使此代码更高效/更短至 运行 吗？

Can i make this code more efficient / shorter to run?

javascript

algorithm

wordle-game

我可以使此代码更高效/更短至运行吗？