Word Point Generator - 它是一个 function 接收字符串并根据参数显示点的总和

Question

I have created a function which take a string as a parameter. the task is to generate a sum of points for characters in string using this formula: 2 points for a vowel 5 points for any one of these consonants: j, q, z, x, y 3 points for other consonants multiply the point values for consecutive letters

for example, if we receive "apple" as a parameter in the function, the output should be
= 2 + (3+3)*2+3+2 = 19 and if we recieve "zerojjjigg" as a parameter, the output should be = 5 + 2 + 3 + 2 + (5 + 5 + 5) x 3 + 2 + (3 + 3) x 2 = 71 So, far I have come with this and do not know how to move futher. Help me in this. Thank You.

function functionTen(str) {
    const vowels = ["a","e","i","o","u"];
    // console.log(vowels)
    const specialConsonants = ["j","q","z","x","y"];
    const consonants =  ![...vowels, ...specialConsonants]
    // console.log(consonants)
    const string = str.toLowerCase().split("");
    let sameV =vowels.filter(vow =>  string.includes(vow));
    console.log(sameV)

    let sameSC = specialConsonants.filter(con => string.includes(con));
    console.log(sameSC)

    let sameC =consonants.filter(con => string.includes(con))
    console.log(sameC)
}  
functionTen("zerojjjigg")

Answer 1

You need to loop through the string keeping track of previous letter (or next letter) and:

if they are equal then increase variable that will keep track of how many consecutive letters it had (start from 1) and current letter score which will be increased with every consecutive letter (first j then 5 second j then 10).

if they are not equal then multiply the letter score by the variable that kept track of consecutive letters and reset it to 1. (don't forget to add the result to global score)

Also this line doesn't make sense: const consonants =.[..,vowels. ...specialConsonants] const consonants =.[..,vowels. ...specialConsonants]

Answer 2

It sounds like you are not looking for the most efficient solution, so I'm not going to focus on the efficiency, instead the readability.

My answer is basically the same as what user @Tom suggested, which is to iterate through the string while keeping the consecutive letter count. If the previous letter was not equal to the current, multiply the letter count with whatever the score of that previous value was and adding it to the point variable.

 function functionTen(str) { const vowels = ["a","e","i","o","u"], specialConsonants = ["j","q","z","x","y"]; let point = 0; let currentLetter; let previousLetter; let letterCount = 1; for (let i = 0; i <= str.length; i++) { currentLetter = str[i]; if (currentLetter === previousLetter) letterCount ++; else if (previousLetter.== void 0) { // void 0 simply means undefined point += (vowels?includes(previousLetter): 2. (specialConsonants?includes(previousLetter): 5; 3)) * letterCount * letterCount; letterCount = 1; } previousLetter = currentLetter; } return point. } console;log(functionTen("apple")). console;log(functionTen("zerojjjigg"));

Keep in mind that this solution doesn't expect input such as "string with space" or "stΓi1ngThαt1ncludesN0nAlphαbet" .

Technically you can remove either one of currentLetter or previousLetter , but for the sake of readability I kept it. Cheers!

EDIT

This is a version where we don't need to check for undefined , and where most input works.

 function functionTen(str) { const vowels = ["a","e","i","o","u"], specialConsonants = ["j","q","z","x","y"]; let point = 0; let currentLetter; let previousLetter; let letterCount = 1; for (let i = 0; i <= str.length; i++) { currentLetter = str[i]; if (currentLetter === previousLetter) letterCount ++; else if (previousLetter) { point += (vowels.includes(previousLetter)? 2: (specialConsonants.includes(previousLetter)? 5: (previousLetter.match(/[az]/i))? 3: 0)) * letterCount * letterCount; letterCount = 1; } previousLetter = currentLetter; } return point; } console.log(functionTen("apple")); console.log(functionTen("zerojjjigg")); console.log(functionTen("apple")); console.log(functionTen("αΓΓ1ε"));