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palindrome in javascript
function palindrome(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]) {
return false;
}
}
return true;
}
Source: stackoverflow.com
javascript palindrome check
function palindrome(str) {
// Step 1. The first part is the same as earlier
var re = /[^A-Za-z0-9]/g; // or var re = /[\W_]/g;
str = str.toLowerCase().replace(re, '');
// Step 2. Create the FOR loop
var len = str.length; // var len = "A man, a plan, a canal. Panama".length = 30
for (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 on
return false; // 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*/
}
return true; // Both parts are strictly equal, it returns true => The string is a palindrome
}
palindrome("A man, a plan, a canal. Panama");
Source: www.freecodecamp.org
javascript palindrome check
function palindrome(str) {
// Step 1. Lowercase the string and use the RegExp to remove unwanted characters from it
var 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 Boolean
return reverseStr === lowRegStr; // "amanaplanacanalpanama" === "amanaplanacanalpanama"? => true
}
palindrome("A man, a plan, a canal. Panama");
Source: www.freecodecamp.org
js palindrome number
isPalindrome = function(x) {
if (x < 0) {
return false;
}
return x === reversedInteger(x, 0);
};
reversedInteger = function(x, reversed) {
if(x<=0) return reversed;
reversed = (reversed * 10) + (x % 10);
x = Math.floor(x / 10);
return reversedInteger(x, reversed);
};
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