有人可以告诉我为什么这种方法对发现回文法无效吗？

Question

Attempting to complete the algorithms on freeCodeCamp. I eventually found an approach that works, but i still don't understand why this method did not work for all cases.

function palindrome(str) {
  var alphaNumericStr = str.replace(/\W/g,""); 
  var lowerCaseAlphaNumericString = alphaNumericStr.toLowerCase();  
  var arr = lowerCaseAlphaNumericString.split(""); 
  arr.reverse();                                   
  var reversedString = arr.join("");               

  if(str === reversedString){
    return true;
  } 
  return false;
 }

palindrome("race car");

Answer 1

You're comparing a string which has been stripped of spaces and converted to lowercase to the original string. Replace your conditional with:

if(lowerCaseAlphaNumericString == reversedString){
    rethrn true;
}
return false;

Here's a little refactor if you're interested:

// ...
var reversedString = arr.join('');
return lowerCaseAlphaNumericString == reversedString;

demo

Answer 2

This is where you are going wrong if(str === reversedString)

Try this:

    if(lowerCaseAlphaNumericString === reversedString) {
        return true;
    } 
    return false;
}

Answer 3

There could be another approach. In this approach, the corner cases are handled separately.

function check_Palindrome(input_str){
    var astr = input_str.toLowerCase().replace(/\W/g,'');
    var acount = 0;
    if(astr==="") {
        console.log("Not Palindrome.");
        return false;
    }
    if ((astr.length) % 2 === 0) {
        acount = (astr.length) / 2;
    } else {
        if (astr.length === 1) {
            console.log("Palindrome.");
            return true;
        } else {
            acount = (astr.length - 1) / 2;
        }
    }
   for (var x = 0; x < acount; x++) {
        if (astr[x] != astr.slice(-1-x)[0]) {
            console.log("Not Palindrome.");
            return false;
        }
    }
    console.log("Palindrome.");
    return true;
}