cómo comprobar si un string es palíndromo en un ejemplo de código javascript

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Ejemplo 1: palíndromo en javascript

functionpalindrome(str)var len = str.length;var mid = Math.floor(len/2);for(var i =0; i < mid; i++)if(str[i]!== str[len -1- i])returnfalse;returntrue;

Ejemplo 2: verificación palíndromo de javascript

functionpalindrome(str)var re =/[W_]/g;var lowRegStr = str.toLowerCase().replace(re,'');var reverseStr = lowRegStr.split('').reverse().join('');return reverseStr === lowRegStr;palindrome("A man, a plan, a canal. Panama");

Ejemplo 3: ¿Cómo se detecta si una palabra es un palíndromo en javascript?

functionpalindrome(str)// Step 1. The first part is the same as earliervar re =/[^A-Za-z0-9]/g;// or var re = /[W_]/g;
 str = str.toLowerCase().replace(re,'');// Step 2. Create the FOR loopvar len = str.length;// var len = "A man, a plan, a canal. Panama".length = 30for(var i =0; i < len/2; i++)if(str[i]!== str[len -1- i])// As long as the characters from each part match, the FOR loop will go onreturnfalse;// When the characters don't match anymore, false is returned and we exit the FOR loop/* Here len/2 = 15
      For each iteration: i = ?    i < len/2    i++    if(str[i] !== str[len - 1 - i])?
      1st iteration:        0        yes         1     if(str[0] !== str[15 - 1 - 0])? => if("a"  !==  "a")? // false
      2nd iteration:        1        yes         2     if(str[1] !== str[15 - 1 - 1])? => if("m"  !==  "m")? // false      
      3rd iteration:        2        yes         3     if(str[2] !== str[15 - 1 - 2])? => if("a"  !==  "a")? // false  
      4th iteration:        3        yes         4     if(str[3] !== str[15 - 1 - 3])? => if("n"  !==  "n")? // false  
      5th iteration:        4        yes         5     if(str[4] !== str[15 - 1 - 4])? => if("a"  !==  "a")? // false
      6th iteration:        5        yes         6     if(str[5] !== str[15 - 1 - 5])? => if("p"  !==  "p")? // false
      7th iteration:        6        yes         7     if(str[6] !== str[15 - 1 - 6])? => if("l"  !==  "l")? // false
      8th iteration:        7        yes         8     if(str[7] !== str[15 - 1 - 7])? => if("a"  !==  "a")? // false
      9th iteration:        8        yes         9     if(str[8] !== str[15 - 1 - 8])? => if("n"  !==  "n")? // false
     10th iteration:        9        yes        10     if(str[9] !== str[15 - 1 - 9])? => if("a"  !==  "a")? // false
     11th iteration:       10        yes        11    if(str[10] !== str[15 - 1 - 10])? => if("c" !==  "c")? // false
     12th iteration:       11        yes        12    if(str[11] !== str[15 - 1 - 11])? => if("a" !==  "a")? // false
     13th iteration:       12        yes        13    if(str[12] !== str[15 - 1 - 12])? => if("n" !==  "n")? // false
     14th iteration:       13        yes        14    if(str[13] !== str[15 - 1 - 13])? => if("a" !==  "a")? // false
     15th iteration:       14        yes        15    if(str[14] !== str[15 - 1 - 14])? => if("l" !==  "l")? // false
     16th iteration:       15        no               
    End of the FOR Loop*/returntrue;// Both parts are strictly equal, it returns true => The string is a palindromepalindrome("A man, a plan, a canal. Panama");

Ejemplo 4: ¿Cómo se detecta si una palabra es un palíndromo en javascript?

functionpalindrome(str)// Step 1. Lowercase the string and use the RegExp to remove unwanted characters from itvar re =/[W_]/g;// or var re = /[^A-Za-z0-9]/g;var lowRegStr = str.toLowerCase().replace(re,'');// str.toLowerCase() = "A man, a plan, a canal. Panama".toLowerCase() = "a man, a plan, a canal. panama"// str.replace(/[W_]/g, '') = "a man, a plan, a canal. panama".replace(/[W_]/g, '') = "amanaplanacanalpanama"// var lowRegStr = "amanaplanacanalpanama";// Step 2. Use the same chaining methods with built-in functions from the previous article 'Three Ways to Reverse a String in JavaScript'var reverseStr = lowRegStr.split('').reverse().join('');// lowRegStr.split('') = "amanaplanacanalpanama".split('') = ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"]// ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"].reverse() = ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"]// ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"].join('') = "amanaplanacanalpanama"// So, "amanaplanacanalpanama".split('').reverse().join('') = "amanaplanacanalpanama";// And, var reverseStr = "amanaplanacanalpanama";// Step 3. Check if reverseStr is strictly equals to lowRegStr and return a Booleanreturn reverseStr === lowRegStr;// "amanaplanacanalpanama" === "amanaplanacanalpanama"? => truepalindrome("A man, a plan, a canal. Panama");

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