JavaScript - 将字符串拆分为单词、识别元音最多的单词的函数 - 使用 For 循环

Question

我检查了一些similar questions但我不认为答案直接适用于我正在寻找的东西。

我试图找到给定字符串中元音最多的单词。我知道如何将字符串拆分成这样的单词:

let Words = string.split("");


到目前为止我已经:

function mostVowels(string) {

  let vowels = ["aeiouAEIOU"];
  let words = string.split(" ");

  //initiate vowel count at 0

  let counter = 0;

  //loop through words

  for (let i = 0; i < words.length; i++) {
    //if there are vowels in the word, add vowels to counter
    if (vowels.includes(i)) {
      counter++
    }
  }
  return counter;
}

console.log(mostVowels("I went to the park today."));

很明显我距离找到解决方案还很远。

现在返回0。

该字符串的前五个单词中的每一个都有 1 个元音，而“today”中有两个元音，因此最终我们希望此函数返回“today”作为元音最多的单词。

此时，我只是想获取字符串中第一个单词的元音计数 - 它应该是 1。

然后我想我将能够比较不同单词的计数以确定哪个计数最大。

Answer 1

将元音定义为字符数组，而不是包含单个字符串的数组。然后，在 words 循环内，将计数器初始化为 0，并迭代单词中的每个字符。如果该字符包含在元音数组中，则递增计数器。在迭代单词结束时，如果该单词的计数器优于迄今为止最好的计数器，则将计数器和单词分配给外部变量，指示迄今为止最好的单词/计数器。在函数结束时，返回最好的单词:

function mostVowels(string) {

  let vowels = [..."aeiouAEIOU"];
  let words = string.split(" ");

  let bestCounter = 0;
  let bestWord;
  for (let i = 0; i < words.length; i++) {
    const word = words[i];
    //initiate vowel count at 0
    let counter = 0;
    for (let i = 0; i < word.length; i++) {
      const char = word[i];
      //if there are vowels in the word, add vowels to counter
      if (vowels.includes(char)) {
        counter++
      }
    }
    // finished iterating through loop,
    // now check to see if it's better than the best word so far:
    if (counter > bestCounter) {
      bestCounter = counter;
      bestWord = word;
    }
  }
  return bestWord;
}

console.log(mostVowels("I went to the park today."));
console.log(mostVowels("I went to the fooaeuiubar park today."));

或者，也许更优雅的是，使用数组方法:

const vowels = [..."aeiouAEIOU"];
const getVowelCount = word => [...word].reduce((a, char) => a + Boolean(vowels.includes(char)), 0);
function mostVowels(string) {
  let bestCounter = 0;
  return string.split(' ')
    .reduce((bestWordSoFar, thisWord) => {
      const thisVowelCount = getVowelCount(thisWord);
      if (thisVowelCount > bestCounter) {
        bestCounter = thisVowelCount;
        return thisWord;
      } else {
        return bestWordSoFar;
      }
    });
}

console.log(mostVowels("I went to the park today."));
console.log(mostVowels("I went to the fooaeuiubar park today."));