# Leetcode - Reverse Integer (with JavaScript)

Today I am going to show how to solve the Leetcode Reverse Integer algorithm problem.

1) I declare the two variables revNum and lastDigit, which represent the reverse and the last digit of giving integer x, respectively. Then, I loop through x and build up the reverse integer one digit at a time.

const reverse = function(x){
let revNum = 0, lastDigit = 0;
}


2) The remainder of a number divided by 10 will give us the last digit.
For example: 123/10 = 12.3 -> the remainder is 3, which is the last digit.

const reverse = function(x){
let revNum = 0, lastDigit = 0;
while (x!==0) {
lastDigit = x % 10;
}
}


3) Then, I remove that digit from the end of x by using parseInt() method.
For example: parseInt(12.3) = 12

const reverse = function(x){
let revNum = 0, lastDigit = 0;
while (x!==0) {
lastDigit = x % 10;
x = parseInt(x/10);
revNum = revNum * 10 + lastDigit; // building up the reverse number
}
}


4) In the problem it was noted that we are dealing with an environment which could only store integers within the 32-bit signed integer range. I therefore check beforehand whether or not appending another digit would cause overflow. If it causes overflow, the loop will break and return 0.

const reverse = function(x){
let revNum = 0, lastDigit = 0;
while (x!==0) {
lastDigit = x % 10;
x = parseInt(x/10);
revNum = revNum * 10 + lastDigit;
if (revNum < Math.pow(-2, 31) || revNum > Math.pow(2, 31) - 1) return 0
}
return revNum
}


### Discussion Hi Urfan, I want to suggest another solution:

const reverse = n => {
const str      = "" + Math.abs(n); // convert absolute value to string
const reversed = str.split("")     // get array of digit characters
.reverse()        // reverse it
.join("");        // join into string again;
const num      = +reversed;        // convert integer
return (n < 0 ? -1 : 1) * num;     // multiple by -1 if needed
}


even shorter:

const reverse = n => (n < 0 ? -1 : 1) * +("" + Math.abs(n)).split.reverse().join;


Cheers.  