检查一个单词是否是纯javascript的等值线

Question

我如何使用函数检查给定单词是否是纯 JavaScript 的等值图。该函数必须返回 true 或 false。

等值线是一个带有重复字符的单词。

我知道这段代码有效，但我需要更好的解决方案。

function isIsogram(word){
    x = false; y = false;
    for(i = 0; i < word.length; i++){
        wordl = word.substring(0,i)
        wordr = word.substring(i)
        x = wordl.includes(word.charAt(i))
        y = wordr.includes(word.charAt(i))
        //console.log(x,wordl,wordr)
    }
    return x&&y
}
isIsogram("thomas");//False
isIsogram("moses"); //True

Answer 1

从字符串中删除重复的字母，然后检查两个长度。如果相同，则为等值线。

function isIsogram(str){
 return str.split('').filter((item, pos, arr)=> arr.indexOf(item) == pos).length == str.length;
}
console.log(isIsogram('thomas'));
console.log(isIsogram('moses'));

Answer 2

一种方法！

function isIsogram(str){
  return !str.match(/([a-z]).*\1/i);
}

Answer 3

这是使用.split() 和.every() 的简单方法：

let isIsogram = (str) => str.split("").every((c, i) => str.indexOf(c) == i);
                            
console.log(isIsogram("thomas"));   /* no repeating letter */
console.log(isIsogram("moses"));    /* s repeat 2 times */
console.log(isIsogram("hello"));    /* l repeat 2 times */
console.log(isIsogram("world"));    /* no repeating letter */
console.log(isIsogram("a b c"));    /* space character repeat 2 times */

文档：

• String.prototype.split()
• String.prototype.indexOf()
• Array.prototype.every()

Answer 4

基于 kishea 的回答：

function isIsogram(sWord)
 {
  for (iCharIndex = 0; iCharIndex < sWord.length; iCharIndex++)
    if (sWord.substring(iCharIndex + 1).includes(sWord.charAt(iCharIndex)))
      return false;
  return true;
 }

如果在当前位置 (substring) 的右侧找到当前位置 (charAt) 的字符 (includes)，则返回 false。否则循环运行到最后并返回true。

Answer 5

const isIsogram = (word) => {
    return new Set(word.toLowerCase()).size === word.length
}

console.log(isIsogram('Thor'));//true
console.log(isIsogram('Loki'));//true
console.log(isIsogram('America'));//false

Answer 6

怎么样：

> function isIsogram(word) {
... var a = word.split('')
... var b = Array.from(new Set(a))
... return a.length === b.length;
... }
undefined
> isIsogram("mesos")
false
> isIsogram("thomas")
true

或者更好（每个字符只检查一次）：

> function isIsogram2(word) {
... for(var i=0,len=word.length-1;i<len;++i) {
..... var c = word[i]
..... if(word.indexOf(c,i+1) !== -1) return false;
..... }
... return true;
... }
undefined
> isIsogram2("mesos")
false
> isIsogram2("thomas")
true

Answer 7

function isIsogram(word){
   return !/(.).*\1|\d/i.test(word)
}

Answer 8

var str=prompt('Enter a string');
    var strlen=str.length;

    for(i=0;i<strlen;i++){
       var stc=str.charAt(i);
       var flag=0;
       for(j=0;j<strlen;j++){
        var stk=str.charAt(j);
          if(stc==stk){
            flag=flag+1;
          }
       }

       if(flag!=1){
        break;
       }

    }

    if(flag!=1){
        alert('It is not an isogram');
    }
    else{
        alert('It is an isogram');
    }

Answer 9

当给定一个单词时，这个函数 if 将单词分成两部分， 那分别是wordl和wordr。 检查两个拆分以在原始单词中包含一个字符。如果 wordl 和 wordr 都包含原始单词中的任何字符。那么这肯定是一个等值线

function isIsogram(word){
    x = false; y = false;
    for(i = 0; i < word.length; i++){
        wordl = word.substring(0,i)
        wordr = word.substring(i)
        x = wordl.includes(word.charAt(i))
        y = wordr.includes(word.charAt(i))
        //console.log(x,wordl,wordr)
    }
    return !x&&y
}
isIsogram("thomas");//True
isIsogram("moses"); //False

Answer 10

const isIsogram = (string) => {
    
    const lowerCased = string.toLowerCase()
    
    const result = lowerCased.split('').every((v,i)=>lowerCased.indexOf(v)===i)
    
    return result

}

console.log(isIsogram('ambidExtRously')) // true
console.log(isIsogram('patteRN')) // false

Answer 11

function isIsogram(word) {

    let uniqueCharacters = new Set(word.split(''));

    uniqueCharacters = Array.from(uniqueCharacters); //get all the unique char

    let charFreq = {}; //e.g {a:2, b:3}

    for (element of uniqueCharacters) {
        charFreq[element] = 0;
    } //set the frequency of each char to zero

    function updateFrequency(element) {
        charFreq[element] += 1;
    }//callback used directly below

    word.split('').forEach(updateFrequency); //for each char encountered, update frequency

    let frequencyOfCharacter = [];

    for (keys in charFreq) {
        frequencyOfCharacter.push(charFreq[keys]);
    }

    function equal(item) {
        return item === frequencyOfCharacter[0];
    }
    //check if all the frequencies are the same, an isogram means all characters occur at the same frequency

    return frequencyOfCharacter.every(equal);
}

console.log(isIsogram('try'), isIsogram('baba'), isIsogram('tests'));