javascript palindrome check

1function palindrome(str) {
2
3    var len = str.length;
4    var mid = Math.floor(len/2);
5
6    for ( var i = 0; i < mid; i++ ) {
7        if (str[i] !== str[len - 1 - i]) {
8            return false;
9        }
10    }
11
12    return true;
13}
14

1function palindrome(str) {
2  var re = /[\W_]/g;
3  var lowRegStr = str.toLowerCase().replace(re, '');
4  var reverseStr = lowRegStr.split('').reverse().join(''); 
5  return reverseStr === lowRegStr;
6}
7palindrome("A man, a plan, a canal. Panama");

1function palindrome(str) {
2 var re = /[^A-Za-z0-9]/g;
3 str = str.toLowerCase().replace(re, '');
4 var len = str.length;
5 for (var i = 0; i < len/2; i++) {
6   if (str[i] !== str[len - 1 - i]) {
7       return false;
8   }
9 }
10 return true;
11}
12palindrome("A man, a plan, a canal. Panama");

1function palindrome(str) {
2 // Step 1. The first part is the same as earlier
3 var re = /[^A-Za-z0-9]/g; // or var re = /[\W_]/g;
4 str = str.toLowerCase().replace(re, '');
5
6 // Step 2. Create the FOR loop
7 var len = str.length; // var len = "A man, a plan, a canal. Panama".length = 30
8 
9 for (var i = 0; i < len/2; i++) {
10   if (str[i] !== str[len - 1 - i]) { // As long as the characters from each part match, the FOR loop will go on
11       return false; // When the characters don't match anymore, false is returned and we exit the FOR loop
12   }
13   /* Here len/2 = 15
14      For each iteration: i = ?    i < len/2    i++    if(str[i] !== str[len - 1 - i])?
15      1st iteration:        0        yes         1     if(str[0] !== str[15 - 1 - 0])? => if("a"  !==  "a")? // false
16      2nd iteration:        1        yes         2     if(str[1] !== str[15 - 1 - 1])? => if("m"  !==  "m")? // false      
17      3rd iteration:        2        yes         3     if(str[2] !== str[15 - 1 - 2])? => if("a"  !==  "a")? // false  
18      4th iteration:        3        yes         4     if(str[3] !== str[15 - 1 - 3])? => if("n"  !==  "n")? // false  
19      5th iteration:        4        yes         5     if(str[4] !== str[15 - 1 - 4])? => if("a"  !==  "a")? // false
20      6th iteration:        5        yes         6     if(str[5] !== str[15 - 1 - 5])? => if("p"  !==  "p")? // false
21      7th iteration:        6        yes         7     if(str[6] !== str[15 - 1 - 6])? => if("l"  !==  "l")? // false
22      8th iteration:        7        yes         8     if(str[7] !== str[15 - 1 - 7])? => if("a"  !==  "a")? // false
23      9th iteration:        8        yes         9     if(str[8] !== str[15 - 1 - 8])? => if("n"  !==  "n")? // false
24     10th iteration:        9        yes        10     if(str[9] !== str[15 - 1 - 9])? => if("a"  !==  "a")? // false
25     11th iteration:       10        yes        11    if(str[10] !== str[15 - 1 - 10])? => if("c" !==  "c")? // false
26     12th iteration:       11        yes        12    if(str[11] !== str[15 - 1 - 11])? => if("a" !==  "a")? // false
27     13th iteration:       12        yes        13    if(str[12] !== str[15 - 1 - 12])? => if("n" !==  "n")? // false
28     14th iteration:       13        yes        14    if(str[13] !== str[15 - 1 - 13])? => if("a" !==  "a")? // false
29     15th iteration:       14        yes        15    if(str[14] !== str[15 - 1 - 14])? => if("l" !==  "l")? // false
30     16th iteration:       15        no               
31    End of the FOR Loop*/
32 }
33 return true; // Both parts are strictly equal, it returns true => The string is a palindrome
34}
35
36palindrome("A man, a plan, a canal. Panama");

1function palindrome(str) {
2  // Step 1. Lowercase the string and use the RegExp to remove unwanted characters from it
3  var re = /[\W_]/g; // or var re = /[^A-Za-z0-9]/g;
4  
5  var lowRegStr = str.toLowerCase().replace(re, '');
6  // str.toLowerCase() = "A man, a plan, a canal. Panama".toLowerCase() = "a man, a plan, a canal. panama"
7  // str.replace(/[\W_]/g, '') = "a man, a plan, a canal. panama".replace(/[\W_]/g, '') = "amanaplanacanalpanama"
8  // var lowRegStr = "amanaplanacanalpanama";
9     
10  // Step 2. Use the same chaining methods with built-in functions from the previous article 'Three Ways to Reverse a String in JavaScript'
11  var reverseStr = lowRegStr.split('').reverse().join(''); 
12  // 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"]
13  // ["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"]
14  // ["a", "m", "a", "n", "a", "p", "l", "a", "n", "a", "c", "a", "n", "a", "l", "p", "a", "n", "a", "m", "a"].join('') = "amanaplanacanalpanama"
15  // So, "amanaplanacanalpanama".split('').reverse().join('') = "amanaplanacanalpanama";
16  // And, var reverseStr = "amanaplanacanalpanama";
17   
18  // Step 3. Check if reverseStr is strictly equals to lowRegStr and return a Boolean
19  return reverseStr === lowRegStr; // "amanaplanacanalpanama" === "amanaplanacanalpanama"? => true
20}
21 
22palindrome("A man, a plan, a canal. Panama");

1const isPalindrome = str => str === str.split('').reverse().join('');
2
3// Examples
4isPalindrome('abc');        // false
5isPalindrom('abcba');       // true