Question

我正在尝试解决Codewars上的Javascript挑战。

isogram是一个没有重复字母，连续或不连续的单词。实现一个函数，确定只包含字母的字符串是否是等值线图。假设空字符串是等值线图。忽略字母案例。

isIsogram( "Dermatoglyphics" ) == true
isIsogram( "aba" ) == false
isIsogram( "moOse" ) == false // -- ignore letter case

我的努力如下：

    function isIsogram(str) {
    var arr = str.split("");
    var seen = {};
    var out = [];
    var length = arr.length;
    var j = 0;
    for(var i = 0; i < length; i++) {
         var item = arr[i].toLowerCase;
         if(seen[item] !== 1) {
               seen[item] = 1;
               out[j++] = item;
         }
    }
    console.log(out.toString.toLowerCase);
    console.log(str.toLowerCase);
    if (out.toString.toLowercase === str.toLowerCase) {
     return true;
    }
    else {
      return false;
    }
}

在代码大战中我的结果

console.log(out.toString.toLowerCase); is undefined

和

的结果

console.log(str.toLowerCase); is [Function: toLowerCase].

这意味着我的解决方案始终评估为false。

我很感激任何帮助，指出我正确的方向或突出我的错误，而不是给我解决方案，以便我可以更有效地学习。谢谢！

Answer 1

这可能是一个更简单的答案。

＆＃13;

function isIsogram(str){

  // Turn all letters of the string to lower case and split it into an array. 

  var letters = str.toLowerCase().split('');
  var checkLetters = [];
  
  /* Check to see if the letter appears in the checkLetters array.
     If the letter is not already in the array it will push the letter into it. */

  letters.forEach(function(letter) {
    if(checkLetters.indexOf(letter) === -1) {
      checkLetters.push(letter);
    }
  });

  /* Now we have two arrays. If the letters array has non-duplicate letters then 
     it will be the same length as the checkLetters array. If not, the checkLetters array
     will be shorter. */

  /* Return true or false depending on whether the lengths of both arrays are equal */
    
  return letters.length === checkLetters.length ? true : false;

}

＆＃13;

Answer 2

toString和toLowerCase是函数。在Javascript中，要执行函数，必须在函数名的末尾添加括号：

out.toString().toLowerCase()
//          ^^            ^^

您需要为所有功能执行此操作：

arr[i].toLowerCase()

str.toLowerCase

out.toString.toLowercase() === str.toLowerCase()

（请注意，在数组上调用.toString()将包含逗号，例如"a,b,c,d,e"。对于此用例可能无关紧要，但只是为了突出显示） < / p>

Answer 3

toString和toLowerCase等是函数

使用：

out.toString().toLowerCase()

但是，对于out，我认为你想做out.join（''）。toLowerCase（）

Answer 4

我认为您可以通过功能方式，使用 hashmap 系统和every方法来改进您的解决方案。

每个方法对数组中存在的每个元素执行一次提供的回调函数，直到找到一个回调返回伪值的文件

所以，你的代码会更干净。

  function isIsogram(str) {
    //Create our hashmap
    var hashmap = {};
    //Return value of 'every' loop
    return str.split("").every(function(elm){
      //If our hashmap get current letter as key
      return hashmap.hasOwnProperty(elm.toLowerCase())
      //Return false, str is not an Isogram
      ? false
      //Otherwise, set letter as key in our hashmap,
      //we can check the next iteration
      : hashmap[elm.toLowerCase()] = true;
    });
  }

console.log(isIsogram('Dermatoglyphics'));
console.log(isIsogram('moOse'));
console.log(isIsogram('aba'));

Answer 5

我知道你已经解决了这个问题，但只是为了变化。通过忽略区分大小写，首先将输入更改为小写。

function isIsogram(str) {
     // Change word to lower case
    str = str.toLowerCase();
      // split the word into letters and store as an array
    var arr = str.split("");
      // get the length so the array can be looped through
    var len = arr.length;
      // an array to contain letters that has been visited
    var seen = []
    for (var i = 0; i < len; i++) {
        // check if letter has not been visited by checking for the index
      if(seen.indexOf(arr[i])<0){
         // if the letter doesn't exist,add it to seen and go to next letter
        seen.push(arr[i]);
      }else{
         // if letter exists, the word is not an isogram return false
        return false
      }
    }
       // the loop is complete but each letter appeared once, the word is an isogram
   return true
}

console.log(isIsogram('Dermatoglyphics'));
console.log(isIsogram('moOse'));
console.log(isIsogram('aba'));

将字符串转换为数组然后再返回时出现Javascript问题

5 个答案: